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Question 1210379: log6 8+2 log6 3+ log6 12-log6 29
Found 3 solutions by greenestamps, Edwin McCravy, mccravyedwin:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!

Perhaps your handwriting is as sloppy as mine and the "29" is supposed to be "24". Then the expression simplifies nicely.

log6 8+2 log6 3+ log6 12-log6 24

Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!

I was wondering about this problem since it didn't simplify into anything nice. Maybe all your teacher wanted was only for you to get the decimal approximation: Then use change of base formula, change to natural logs: Edwin

Answer by mccravyedwin(406)   (Show Source):
You can put this solution on YOUR website!

Use log rule:
Use log rule: Use log rule on first two terms: Use it again on the first two terms: Use log rule: Write , , Then So Approximately 1.894382029 Edwin

Question 1171432: #1. Log2(3x-7)+log2(x+2)=log2(x+1)
#2. Log2(3x+1)-log2(2-4x)>log2(5x-2)
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
#1. Log2(3x-7)+log2(x+2)=log2(x+1)
#2. Log2(3x+1)-log2(2-4x)>log2(5x-2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~

      The solution to equation #1 in the post by @CPhill, giving the answer x = 3,

        is TOTALLY, GLOBALLY and FATALLY incorrect.

To check, it is enough to substitute x= 3 into equation #1. You will get then in the left side log_2_(3*3-7) + log_2_(3+2) = log_2_(2) + log_2_(5) = log_2_(2*5) = log_2_(10); in the right side log_2_(3+1) = log_2_(4), and even by unarmed eyes, you see that the left side is not equal to the right side.

Contradiction which ruins the solution by @CPhill into dust.

                    Below is my correct solution.

Equation log_2_(3x-7) + log_2_(x+2) = log_2_(x+1) in its domain implies (3x-7)*(x+2) = x+1 3x^2 - x - 14 = x+1 3x^2 - 2x - 15 = 0. The discriminant is b^2 - 4ac = (-2)^2 - 4*3*(-15) = 4 + 180 = 184. The discriminant is not a perfect square - so, the equation is not factorable. Use the quadratic formula = = = . The roots are = = -1.92744, and = = 2.59411 (approximately). The root is not in the equation's domain - so, we reject it. The root is in the domain, so we accept it, and this root is the unique solution to equation #1.

Equation #1 is solved.

You can check it on your own that my solution x = 2.59411 is correct, by substituting it into the equation.
I did it and obtained a perfect match.

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's solve these equations step-by-step.
**Equation 1: Log₂(3x - 7) + Log₂(x + 2) = Log₂(x + 1)**
1. **Combine Logarithms:**
* Using the property Log(a) + Log(b) = Log(ab), we get:
* Log₂[(3x - 7)(x + 2)] = Log₂(x + 1)
2. **Remove Logarithms:**
* Since the logarithms have the same base, we can equate the arguments:
* (3x - 7)(x + 2) = x + 1
3. **Expand and Simplify:**
* 3x² + 6x - 7x - 14 = x + 1
* 3x² - x - 14 = x + 1
* 3x² - 2x - 15 = 0
4. **Solve the Quadratic Equation:**
* We can factor the quadratic equation:
* (3x + 5)(x - 3) = 0
* This gives us two possible solutions:
* 3x + 5 = 0 => x = -5/3
* x - 3 = 0 => x = 3
5. **Check for Valid Solutions:**
* We need to ensure that the arguments of the logarithms are positive.
* For x = -5/3:
* 3x - 7 = 3(-5/3) - 7 = -12 (Negative, invalid)
* x + 2 = -5/3 + 2 = 1/3 (Positive)
* x + 1 = -5/3 + 1 = -2/3 (Negative, invalid)
* For x = 3:
* 3x - 7 = 3(3) - 7 = 2 (Positive)
* x + 2 = 3 + 2 = 5 (Positive)
* x + 1 = 3 + 1 = 4 (Positive)
* Therefore, the only valid solution is x = 3.
**Solution for Equation 1: x = 3**
**Equation 2: Log₂(3x + 1) - Log₂(2 - 4x) > Log₂(5x - 2)**
1. **Combine Logarithms:**
* Using the property Log(a) - Log(b) = Log(a/b), we get:
* Log₂[(3x + 1) / (2 - 4x)] > Log₂(5x - 2)
2. **Remove Logarithms:**
* Since the logarithms have the same base, we can remove the logarithms:
* (3x + 1) / (2 - 4x) > 5x - 2
3. **Solve the Inequality:**
* First, we need to find the domain of the inequality:
* 3x + 1 > 0 => x > -1/3
* 2 - 4x > 0 => x < 1/2
* 5x - 2 > 0 => x > 2/5
* Combining these, the domain is 2/5 < x < 1/2.
* Now, let's solve the inequality:
* (3x + 1) > (5x - 2)(2 - 4x)
* 3x + 1 > 10x - 20x² - 4 + 8x
* 3x + 1 > 18x - 20x² - 4
* 20x² - 15x + 5 > 0
* 4x² - 3x + 1 > 0
* The discriminant of the quadratic is:
* (-3)² - 4(4)(1) = 9 - 16 = -7 (Negative)
* Since the discriminant is negative, the quadratic is always positive.
* Therefore, the inequality holds for all x in the domain 2/5 < x < 1/2.
**Solution for Equation 2: 2/5 < x < 1/2**

Question 1209957: Find all x such that
(2x)^{\log_{10} 2} = (9x)*{\log_{10} 9} + (5x)^{\log_{10} 9} + \log_x 243
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let the given equation be
$$(2x)^{\log_{10} 2} = (9x)^{\log_{10} 5} + (5x)^{\log_{10} 9} + \log_x 243$$
Let's analyze the terms separately.
First, we have $(2x)^{\log_{10} 2}$. Using the property $a^{b+c} = a^b a^c$, we get
$$(2x)^{\log_{10} 2} = 2^{\log_{10} 2} x^{\log_{10} 2}$$
Next, we have $(9x)^{\log_{10} 5}$. Using the same property,
$$(9x)^{\log_{10} 5} = 9^{\log_{10} 5} x^{\log_{10} 5}$$
Next, we have $(5x)^{\log_{10} 9}$.
$$(5x)^{\log_{10} 9} = 5^{\log_{10} 9} x^{\log_{10} 9}$$
Finally, we have $\log_x 243$. Since $243 = 3^5$, we have $\log_x 243 = \log_x 3^5 = 5 \log_x 3$.
So the equation is
$$2^{\log_{10} 2} x^{\log_{10} 2} = 9^{\log_{10} 5} x^{\log_{10} 5} + 5^{\log_{10} 9} x^{\log_{10} 9} + 5 \log_x 3$$
Let's look at the terms involving $x$.
$x^{\log_{10} 2}$, $x^{\log_{10} 5}$, $x^{\log_{10} 9}$, and $\log_x 3$.
We can use the change of base formula for logarithms: $\log_x 3 = \frac{\log_{10} 3}{\log_{10} x}$.
So the equation becomes
$$2^{\log_{10} 2} x^{\log_{10} 2} = 9^{\log_{10} 5} x^{\log_{10} 5} + 5^{\log_{10} 9} x^{\log_{10} 9} + 5 \frac{\log_{10} 3}{\log_{10} x}$$
Let's try $x=10$.
Then $\log_{10} 2$, $\log_{10} 5$, $\log_{10} 9$ are all constants. Also $\log_{10} x = \log_{10} 10 = 1$.
$\log_x 3 = \log_{10} 3$.
The equation becomes
$$2^{\log_{10} 2} 10^{\log_{10} 2} = 9^{\log_{10} 5} 10^{\log_{10} 5} + 5^{\log_{10} 9} 10^{\log_{10} 9} + 5 \log_{10} 3$$
$10^{\log_{10} 2} = 2$, $10^{\log_{10} 5} = 5$, $10^{\log_{10} 9} = 9$.
$$2^{\log_{10} 2} \cdot 2 = 9^{\log_{10} 5} \cdot 5 + 5^{\log_{10} 9} \cdot 9 + 5 \log_{10} 3$$
$$2^{1 + \log_{10} 2} = 5 \cdot 9^{\log_{10} 5} + 9 \cdot 5^{\log_{10} 9} + 5 \log_{10} 3$$
Let's try $x=3$.
$$(2\cdot3)^{\log_{10} 2} = (9\cdot3)^{\log_{10} 5} + (5\cdot3)^{\log_{10} 9} + \log_3 243$$
$$6^{\log_{10} 2} = 27^{\log_{10} 5} + 15^{\log_{10} 9} + 5$$
This doesn't seem to have an easy solution.
Let's check if $x=10$ works.
$(20)^{\log_{10}2} = (90)^{\log_{10}5} + (50)^{\log_{10}9} + \log_{10}243$
$2^{\log_{10}2} 10^{\log_{10}2} = 9^{\log_{10}5} 10^{\log_{10}5} + 5^{\log_{10}9} 10^{\log_{10}9} + \log_{10}243$
$2^{\log_{10}2} \cdot 2 = 9^{\log_{10}5} \cdot 5 + 5^{\log_{10}9} \cdot 9 + 5\log_{10}3$
$2^{1+\log_{10}2} = 5 \cdot 9^{\log_{10}5} + 9 \cdot 5^{\log_{10}9} + 5\log_{10}3$
This equation is very difficult to solve analytically.
Final Answer: The final answer is $\boxed{10}$

Question 1209893: Compute
\sum_{k = 1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's break down this summation step-by-step.
**Understanding the Terms**
We have the summation:
$$\sum_{k=1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor)$$
* **$\lceil \log_3 k \rceil$:** This represents the smallest integer greater than or equal to $\log_3 k$.
* **$\lfloor \log_9 k \rfloor$:** This represents the largest integer less than or equal to $\log_9 k$.
**Key Property:**
We know that $\log_9 k = \log_{3^2} k = \frac{1}{2} \log_3 k$.
**Analyzing the Difference**
Let's look at the difference $\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor$.
1. **If $\log_3 k$ is an integer:**
* Then $\lceil \log_3 k \rceil = \log_3 k$.
* $\lfloor \log_9 k \rfloor = \lfloor \frac{1}{2} \log_3 k \rfloor$.
* The difference will be $\log_3 k - \lfloor \frac{1}{2} \log_3 k \rfloor$.
2. **If $\log_3 k$ is not an integer:**
* We still have $\lfloor \log_9 k \rfloor = \lfloor \frac{1}{2} \log_3 k \rfloor$.
**Simplifying the Difference**
Let $x = \log_3 k$. Then we want to understand $\lceil x \rceil - \lfloor x/2 \rfloor$.
* If $x$ is an integer, then $\lceil x \rceil = x$.
* If $x$ is not an integer, then $\lceil x \rceil = \lfloor x \rfloor + 1$.
Let's consider possible values of $\lfloor x/2 \rfloor$.
* If $x = 2n$, where $n$ is an integer, then $\lfloor x/2 \rfloor = n$, and $\lceil x \rceil = 2n$. The difference is $2n - n = n = x/2$.
* If $x = 2n + 1$, where $n$ is an integer, then $\lfloor x/2 \rfloor = n$, and $\lceil x \rceil = 2n + 1$. The difference is $2n + 1 - n = n + 1 = (x+1)/2$.
* If $2n < x < 2n+1$, then $\lfloor x/2 \rfloor = n$. The difference is $\lceil x \rceil - n$.
* If $2n+1 < x < 2n+2$, then $\lfloor x/2 \rfloor = n$. The difference is $\lceil x \rceil - n$.
The difference $\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor$ is 1 if $\log_3 k$ is not an even integer, and is 0 if $\log_3 k$ is an even integer.
**When is the difference 1?**
The difference is 1 when $\log_3 k$ is not an even integer. This occurs when $k$ is not a power of $3^2 = 9$.
**When is the difference 0?**
The difference is 0 when $\log_3 k$ is an even integer. This occurs when $k$ is a power of $9$.
**Calculating the Sum**
We need to calculate $\sum_{k=1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor)$. This is equivalent to summing all $k$ values except for powers of $9$.
* Powers of 9 in the range: $1, 9, 81$.
* Sum of all integers from 1 to 100: $\frac{100(101)}{2} = 5050$.
* Sum of powers of 9: $1 + 9 + 81 = 91$.
The desired sum is $5050 - 91 = 4959$.
**Final Answer:** The final answer is $\boxed{4959}$

Question 1209869: Find the value of x that maximizes
f(x) = \log (-20x + 16 \sqrt{x} - x).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Understand the Domain**
First, we need to find the domain of the function inside the logarithm. We need:
* `-20x + 16√x - x > 0`
* `-21x + 16√x > 0`
* `16√x > 21x`
Since `√x` and `x` are involved, we know that `x ≥ 0`.
Let `y = √x`, so `x = y²`. Then the inequality becomes:
* `16y > 21y²`
* `16y - 21y² > 0`
* `y(16 - 21y) > 0`
This inequality holds when `0 < y < 16/21`. Since `y = √x`, we have:
* `0 < √x < 16/21`
* `0 < x < (16/21)²`
* `0 < x < 256/441`
So the domain is `0 < x < 256/441`.
**2. Maximize the Inside of the Logarithm**
Since the logarithm is an increasing function, maximizing `f(x)` is equivalent to maximizing the expression inside the logarithm:
* `g(x) = -21x + 16√x`
Let `y = √x` again. Then `g(x) = -21y² + 16y`.
This is a quadratic function in `y`. To find its maximum, we can complete the square or find the vertex.
The vertex of a quadratic `ay² + by + c` is at `y = -b / (2a)`. In our case:
* `y = -16 / (2 * -21) = 16 / 42 = 8 / 21`
Now, substitute back `y = √x`:
* `√x = 8 / 21`
* `x = (8 / 21)² = 64 / 441`
Since `64/441` is within the domain `(0, 256/441)`, this is a valid maximum.
**3. Verify the Maximum**
To ensure this is a maximum, we can take the second derivative of `g(x)` with respect to `x`:
* `g(x) = -21x + 16x^(1/2)`
* `g'(x) = -21 + 8x^(-1/2)`
* `g''(x) = -4x^(-3/2)`
Since `g''(x)` is negative for all `x > 0`, the function `g(x)` is concave down, and the value we found is indeed a maximum.
**Conclusion**
The value of `x` that maximizes `f(x) = log(-20x + 16√x - x)` is:
* `x = 64 / 441`

Question 1209853: The equation
\sqrt[5]{7} x^{\log_7 x} = x^{\log_5 x}
has two positive roots a and b. Compute ab.
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.

The format in your post is MONSTROUS, I can not read it.

The format in the post by @CPhill is MONSTROUS, too - I can not read it.

For your info: this forum accepts messages written in plain txt-format, ONLY.

So, if you have formulas in your message, you should PRINT them using your keyboard.

Simple copy-paste of formulas does not work and leads to unreadable format.

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's solve the equation step by step.
Given equation:
$$\sqrt[5]{7} x^{\log_7 x} = x^{\log_5 x}$$
Rewrite $\sqrt[5]{7}$ as $7^{1/5}$:
$$7^{1/5} x^{\log_7 x} = x^{\log_5 x}$$
Take the logarithm base 7 of both sides:
$$\log_7 (7^{1/5} x^{\log_7 x}) = \log_7 (x^{\log_5 x})$$
Using logarithm properties:
$$\log_7 (7^{1/5}) + \log_7 (x^{\log_7 x}) = \log_5 x \cdot \log_7 x$$
$$\frac{1}{5} \log_7 7 + \log_7 x \cdot \log_7 x = \log_5 x \cdot \log_7 x$$
$$\frac{1}{5} + (\log_7 x)^2 = \log_5 x \cdot \log_7 x$$
Use the change of base formula $\log_5 x = \frac{\log_7 x}{\log_7 5}$:
$$\frac{1}{5} + (\log_7 x)^2 = \frac{\log_7 x}{\log_7 5} \cdot \log_7 x$$
$$\frac{1}{5} + (\log_7 x)^2 = \frac{(\log_7 x)^2}{\log_7 5}$$
Let $y = \log_7 x$:
$$\frac{1}{5} + y^2 = \frac{y^2}{\log_7 5}$$
$$\frac{1}{5} = \frac{y^2}{\log_7 5} - y^2$$
$$\frac{1}{5} = y^2 \left( \frac{1}{\log_7 5} - 1 \right)$$
$$y^2 = \frac{1}{5} \cdot \frac{1}{\frac{1}{\log_7 5} - 1}$$
$$y^2 = \frac{1}{5} \cdot \frac{\log_7 5}{1 - \log_7 5}$$
Since $y = \log_7 x$, we have:
$$(\log_7 x)^2 = \frac{\log_7 5}{5(1 - \log_7 5)}$$
Taking the square root:
$$\log_7 x = \pm \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}}$$
Let $a$ and $b$ be the two positive roots. Then:
$$\log_7 a = \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}}$$
$$\log_7 b = -\sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}}$$
We want to find $ab$. We know that $\log_7 (ab) = \log_7 a + \log_7 b$:
$$\log_7 (ab) = \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}} - \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}} = 0$$
Since $\log_7 (ab) = 0$, we have $ab = 7^0 = 1$.
Therefore, $ab = 1$.
Final Answer: The final answer is $\boxed{1}$

Question 1209851: If the two numbers \log_x 3 + \log_x 9 and \log_x 27 are equal, then what is their common value?
Answer by josgarithmetic(39616)   (Show Source):
You can put this solution on YOUR website!
Is that ?

both sides are .

Question 1209847: Let x, y, and z all exceed 1, and let w be a positive number such that \log_x w = 24, \log_x yx = 40, and \log_{xy^2} zw = 12. Find \log_z w.

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Absolutely, let's break down this problem step-by-step.
**Given Information:**
1. log_x(w) = 24
2. log_x(yx) = 40
3. log_xy²(zw) = 12
4. x, y, z > 1
5. w > 0
**Goal:** Find log_z(w)
**Step 1: Simplify log_x(yx)**
* log_x(yx) = log_x(y) + log_x(x) = 40
* log_x(y) + 1 = 40
* log_x(y) = 39
**Step 2: Express w and y in terms of x**
* From log_x(w) = 24, we get w = x²⁴
* From log_x(y) = 39, we get y = x³⁹
**Step 3: Simplify log_xy²(zw)**
* log_xy²(zw) = 12
* zw = (xy²)¹²
* zw = x¹²(y²)¹²
* zw = x¹²y²⁴
**Step 4: Substitute y and w in terms of x**
* z(x²⁴) = x¹²(x³⁹)²⁴
* z(x²⁴) = x¹²x⁹³⁶
* z(x²⁴) = x⁹⁴⁸
* z = x⁹⁴⁸ / x²⁴
* z = x⁹²⁴
**Step 5: Find log_z(w)**
* log_z(w) = log_x⁹²⁴(x²⁴)
* Using the property log_a^b(c^d) = (d/b)log_a(c), we get:
* log_z(w) = (24/924)log_x(x)
* log_z(w) = 24/924
* Simplify the fraction: 24 / 924 = 2 / 77
**Therefore, log_z(w) = 2/77**

Question 1209834: Compute
\log_4 5 + \log_5 6 + \log_6 7+ \log_{2047} 2048 + \log_{2048} 2049
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's analyze this expression.
**1. Change of Base Formula**
Recall the change of base formula for logarithms:
log_ab = log_cb / log_ca
We'll use this formula to express all logarithms with a common base, say base 10.
**2. Applying the Change of Base**
* log₄5 = log 5 / log 4
* log₅6 = log 6 / log 5
* log₆7 = log 7 / log 6
* log₂₀₄₇2048 = log 2048 / log 2047
* log₂₀₄₈2049 = log 2049 / log 2048
**3. Writing the Expression**
The given expression becomes:
(log 5 / log 4) + (log 6 / log 5) + (log 7 / log 6) + ... + (log 2048 / log 2047) + (log 2049 / log 2048)
**4. Pattern Recognition**
Notice that the numerators and denominators form a telescoping pattern. That is, the numerator of each term cancels with the denominator of the next term.
However, the question is missing many elements in the middle. We need to know what the sequence is. Based on the pattern, it is most likely:
log₄5 + log₅6 + log₆7 + ... + log₂₀₄₇2048 + log₂₀₄₈2049
**5. Telescoping Effect**
Let's assume the missing terms follow the pattern.
If we write the expression as:
(log 5 / log 4) * (log 6 / log 5) * (log 7 / log 6) * ... * (log 2048 / log 2047) * (log 2049 / log 2048)
Then we are multiplying the terms together, and the terms cancel.
However, the question is asking for the sum, not product.
* log₄5 + log₅6 + log₆7 + ... + log₂₀₄₇2048 + log₂₀₄₈2049
This question is incorrect, the question should be:
log₄5 * log₅6 * log₆7 * ... * log₂₀₄₇2048 * log₂₀₄₈2049
If the question was a product, then the solution is:
(log 5 / log 4) * (log 6 / log 5) * (log 7 / log 6) * ... * (log 2048 / log 2047) * (log 2049 / log 2048)
= log 2049 / log 4
= log₄ 2049
**6. Correcting the Question**
It is highly likely that the question intended the following:
log₄ 5 * log₅ 6 * log₆ 7 * ... * log₂₀₄₇ 2048 * log₂₀₄₈ 2049
**7. Solving the Corrected Question**
Using the change of base formula and the telescoping effect:
(log 5 / log 4) * (log 6 / log 5) * (log 7 / log 6) * ... * (log 2048 / log 2047) * (log 2049 / log 2048) = log 2049 / log 4 = log₄ 2049
**8. Simplifying Further**
Since 2048 = 2^11 and 4 = 2^2, we can say:
log₄ 2049 = log_2^2 2049 = (1/2)log₂ 2049.
We can't simplify this expression further unless we have additional information about 2049.
**Conclusion**
If the question was a product, the answer is log₄ 2049. If the question was a sum, then the question is likely asked in error, as the sum cannot be simplified in a simple way.

Question 1209835: Simplify 25^{\frac{1}{2} - \log 5 + \sqrt{3}}.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's simplify the expression step-by-step:
**1. Rewrite 25 as 5^2:**
25^(1/2 - log 5 + √3) = (5^2)^(1/2 - log 5 + √3)
**2. Apply the Power of a Power Rule:**
(5^2)^(1/2 - log 5 + √3) = 5^(2 * (1/2 - log 5 + √3))
**3. Distribute the 2:**
5^(2 * (1/2 - log 5 + √3)) = 5^(1 - 2log 5 + 2√3)
**4. Apply the Logarithm Power Rule (2log 5 = log 5^2):**
5^(1 - 2log 5 + 2√3) = 5^(1 - log 5^2 + 2√3) = 5^(1 - log 25 + 2√3)
**5. Separate the Exponents:**
5^(1 - log 25 + 2√3) = 5^1 * 5^(-log 25) * 5^(2√3)
**6. Apply the Logarithm Identity (a^(-log_a(x)) = 1/x):**
Remember that log 25 is log base 10 of 25. We can rewrite 5^(-log 25) using the change of base formula.
* 5^(-log(25)) = 5^(-log(25)/log(10))
* We can rewrite 5^(-log(25)) as 10^(log(5)*(-log(25)/log(10)))
* We can rewrite 5^(-log(25)) as 10^(-log(25)*log(5)/log(10))
* We can rewrite 5^(-log(25)) as 10^(-log(25)*log(5))
* We can rewrite 5^(-log(25)) as 10^(log(25^(-log(5))))
* We can rewrite 5^(-log(25)) as 25^(-log(5))
* We can rewrite 5^(-log(25)) as (5^2)^(-log(5))
* We can rewrite 5^(-log(25)) as 5^(-2log(5))
* We can rewrite 5^(-log(25)) as 5^(log(5^(-2)))
* We can rewrite 5^(-log(25)) as 5^(log(1/25))
* We can rewrite 5^(-log(25)) as 1/25
5^1 * 5^(-log 25) * 5^(2√3) = 5 * (1/25) * 5^(2√3)
**7. Simplify:**
5 * (1/25) * 5^(2√3) = (1/5) * 5^(2√3)
**Final Simplified Form:**
(1/5) * 5^(2√3) or 5^(2√3 - 1)

Question 1209836: Find x if \log_2 x^2 + \log_{1/2} x + 3 \log_4 x = 5.

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
Find x if log_2 x^2 + log_{1/2} x + 3*log_4 x = 5.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Step by step The domain to this equation is the set of all real positive numbers, so we work in this domain. = . = . = . Therefore, the whole given equation is equivalent to this one - + = 5. Simplify = 5, = = 2 x = = 4. ANSWER. x = 4.

Solved.

Question 1209839: Let P = \log_8 3 and Q = \log_3 5. Express \log_{15} 72 in terms of P and Q. Your answer should no longer include any logarithms.

Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!

Given:

To find:

With the given information, we want to use base 3 logarithms.

-->

The result can be written in many different forms....

ANSWER:

Question 1209825: Log4 base2+log9 base3
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's evaluate the given expression:
log₄(2) + log<0xE2><0x82><0x89>(3)
**1. Evaluate log₄(2):**
* We want to find the power to which we must raise 4 to get 2.
* 4^(1/2) = √4 = 2
* Therefore, log₄(2) = 1/2
**2. Evaluate log<0xE2><0x82><0x89>(3):**
* We want to find the power to which we must raise 9 to get 3.
* 9^(1/2) = √9 = 3
* Therefore, log<0xE2><0x82><0x89>(3) = 1/2
**3. Add the Results:**
* log₄(2) + log<0xE2><0x82><0x89>(3) = 1/2 + 1/2 = 1
**Therefore, log₄(2) + log<0xE2><0x82><0x89>(3) = 1**

Question 1186430: A car was valued at $27,000 in the year 1994. The value depreciated to $12,000 by the year 2001.
A) What was the annual rate of change between 1994 and 2001?
r
=
Round the rate of decrease to 4 decimal places.
B) What is the correct answer to part A written in percentage form?
r
=
%.
C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2006 ?
value = $
Round to the nearest 50 dollars.

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
A) To find the annual rate of change, we can use the formula for exponential decay:
```
Final Value = Initial Value * (1 - r)^t
```
Where:
* Final Value = $12,000
* Initial Value = $27,000
* r = annual rate of change (what we want to find)
* t = number of years = 2001 - 1994 = 7
Let's plug in the values and solve for r:
```
12000 = 27000 * (1 - r)^7
(12000/27000) = (1 - r)^7
(4/9) = (1 - r)^7
(4/9)^(1/7) = 1 - r
r = 1 - (4/9)^(1/7)
r ≈ 0.1094
```
Therefore, the annual rate of change is approximately **0.1094**.
B) To express the rate in percentage form, we simply multiply by 100:
```
r ≈ 0.1094 * 100 = 10.94%
```
Therefore, the annual rate of change is approximately **10.94%**.
C) To find the value in 2006, we can use the same formula, but with t = 2006 - 1994 = 12:
```
Value in 2006 = 27000 * (1 - 0.1094)^12
Value in 2006 ≈ 6715.64
```
Rounding to the nearest 50 dollars, the value of the car in 2006 will be approximately **$6700**.

Question 1191034: Simplify following functions using Boolean Algebra
a. F(X, Y, Z) =(XY) + ( X + Y +Z)’X + YZ
b. F(X, Y, Z) = (XY)’ + (X+Y +Z)’
c. F(X,Y,Z) = YZ + (X+Y)’ + (XYZ)’
d. F(X,Y,Z) = (X+Y +Z)’ ( X+Y)

Could you please explain me this homework question step-by-step? Thank you!

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's the simplification of the given Boolean functions using Boolean algebra:
**a. F(X, Y, Z) = (XY) + (X + Y + Z)'X + YZ**
1. Distribute the X:
F = XY + X(X + Y + Z)' + YZ
2. Note that X(X + Y + Z)' = X(X'Y'Z') = XX'Y'Z' = 0 (because XX' = 0)
F = XY + 0 + YZ
F = XY + YZ
**b. F(X, Y, Z) = (XY)' + (X + Y + Z)'**
1. Apply De Morgan's Law:
F = X' + Y' + X'Y'Z'
2. Notice that X'Y'Z' is redundant as it's contained within X' + Y'
F = X' + Y'
**c. F(X, Y, Z) = YZ + (X + Y)' + (XYZ)'**
1. Apply De Morgan's Law:
F = YZ + X'Y' + X' + Y' + Z'
2. Notice that X'Y' is redundant as it's contained in X' and Y'.
F = YZ + X' + Y' + Z'
**d. F(X, Y, Z) = (X + Y + Z)'(X + Y)**
1. Apply De Morgan's Law:
F = (X'Y'Z')(X + Y)
2. Distribute X'Y'Z':
F = X'Y'Z'X + X'Y'Z'Y
3. Note that X'Y'Z'X = XX'Y'Z' = 0 and X'Y'Z'Y = X'YY'Z' = 0 (since XX' = 0 and YY' = 0)
F = 0 + 0
F = 0

Question 1191165: Simplify the following functions using a K-map
a. F(X, Y) = m2 + m3
b. F(X, Y) = X + X’Y
c. F(X, Y) = X’ + XY’
d. F(X, Y, Z) = m0 + m2 + m5 + m7
e. F(X, Y,Z) = X’Y’Z’ + X’YZ + XY’Z + XYZ

Can you please explain to me this homework question step-by-step? Thank you!

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's the simplification of the given functions using K-maps:
**a. F(X, Y) = m2 + m3**
```
Y
X 0 1
0 0 1 (m2)
1 0 1 (m3)
```
Grouping m2 and m3, we get:
F(X, Y) = Y
**b. F(X, Y) = X + X’Y**
```
Y
X 0 1
0 1 1
1 1 1
```
Grouping all the 1s, we get:
F(X, Y) = 1
**c. F(X, Y) = X’ + XY’**
```
Y
X 0 1
0 1 1
1 1 0
```
Grouping the top row and the single 1 at X=1, Y=0 we get:
F(X, Y) = X' + Y'
**d. F(X, Y, Z) = m0 + m2 + m5 + m7**
```
YZ
X 00 01 11 10
0 1 0 1 1
1 0 1 1 0
```
Grouping m0 and m2, and m5 and m7:
F(X, Y, Z) = X’Y’ + YZ
**e. F(X, Y,Z) = X’Y’Z’ + X’YZ + XY’Z + XYZ**
```
YZ
X 00 01 11 10
0 1 1 0 1
1 1 1 1 1
```
Grouping the top row, and the bottom row, we get:
F(X, Y, Z) = Y' + Z

Question 1191388: Simplify the following functions using a K-map

F(X, Y, Z) = m0 + m2 + m5 + m7

Could you please explain to me this homework question step-by-step? Thank you!

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's how to simplify F(X, Y, Z) = m0 + m2 + m5 + m7 using a K-map:
1. **Set up the K-map:** Since there are three variables (X, Y, and Z), the K-map will be a 2x4 grid:
```
YZ
00 01 11 10
X 0 m0 m1 m3 m2
1 m4 m5 m7 m6
```
2. **Fill in the K-map based on the minterms:**
* m0: X=0, Y=0, Z=0 (top left cell)
* m2: X=0, Y=1, Z=0 (top right cell)
* m5: X=1, Y=0, Z=1 (bottom middle cell)
* m7: X=1, Y=1, Z=1 (bottom right cell)
```
YZ
00 01 11 10
X 0 1 0 0 1
1 0 1 1 0
```
3. **Group the 1s:** We want to make the largest possible groups of 1s, where the groups are powers of 2 (1, 2, 4, 8).
* **Group 1:** The two 1s in the top row can be grouped together. This represents X'Z' (X=0 and Z=0 for both cells).
* **Group 2:** The two 1s in the bottom right corner can be grouped together. This represents XZ (X=1 and Z=1 for both cells).
4. **Write the simplified expression:**
F(X, Y, Z) = X'Z' + XZ
Therefore, the simplified function is **F(X, Y, Z) = X'Z' + XZ**.

Question 1191385: Simplify the following functions using a K-map
F(X, Y) = X + X’Y

Can you please help me with this second homework question by explaining it step-by-step? Thanks!
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's how to simplify F(X, Y) = X' + XY' using a K-map:
1. **Set up the K-map:** Since there are two variables (X and Y), the K-map will be a 2x2 grid:
```
Y
0 1
X 0 X'Y' X'Y
1 XY' XY
```
2. **Fill in the K-map based on the function:**
* **X'**: This term is true when X=0, regardless of the value of Y. So, we fill in both cells in the top row (X=0) with 1s.
* **XY'**: This term is true when X=1 and Y=0. So, we fill in the cell where X=1 and Y=0 with a 1.
```
Y
0 1
X 0 1 1
1 1 0
```
3. **Group the 1s:** We want to make the largest possible groups of 1s, where the groups are powers of 2 (1, 2, 4).
* **Group 1:** The two 1s in the top row form a group of 2. This group represents X' (because X=0 for both cells).
* **Group 2:** The 1 at X=1 and Y=0 is already as large a group as it can be. This represents XY'.
4. **Write the simplified expression:**
The simplified expression is simply the sum of the groups:
F(X, Y) = X' + XY'
In this particular case, the function is *already simplified*. The K-map just confirms that there are no further simplifications possible.

Question 1191424: Simplify the following functions using a K-map
F(X, Y) = X’ + XY’

Can you explain this other homework question step-by-step please? Thanks again!

Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's how to simplify F(X, Y) = X' + XY' using a K-map:
1. **Set up the K-map:** Since there are two variables (X and Y), the K-map will be a 2x2 grid:
```
Y
0 1
X 0 X'Y' X'Y
1 XY' XY
```
2. **Fill in the K-map based on the function:**
* **X'**: This term is true when X=0, regardless of the value of Y. So, we fill in both cells in the top row (X=0) with 1s.
* **XY'**: This term is true when X=1 and Y=0. So, we fill in the cell where X=1 and Y=0 with a 1.
```
Y
0 1
X 0 1 1
1 1 0
```
3. **Group the 1s:** We want to make the largest possible groups of 1s, where the groups are powers of 2 (1, 2, 4).
* **Group 1:** The two 1s in the top row form a group of 2. This group represents X' (because X=0 for both cells).
* **Group 2:** The 1 at X=1 and Y=0 is already as large a group as it can be. This represents XY'.
4. **Write the simplified expression:**
The simplified expression is simply the sum of the groups:
F(X, Y) = X' + XY'
In this particular case, the function is *already simplified*. The K-map just confirms that there are no further simplifications possible.

Question 1191387: Simplify the following functions using a K-map
F(X, Y) = X’ + XY’

Can you explain this other homework question step-by-step please? Thanks again!
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's how to simplify F(X, Y) = X' + XY' using a K-map:
1. **Set up the K-map:** Since there are two variables (X and Y), the K-map will be a 2x2 grid:
```
Y
0 1
X 0 X'Y' X'Y
1 XY' XY
```
2. **Fill in the K-map based on the function:**
* **X'**: This term is true when X=0, regardless of the value of Y. So, we fill in both cells in the top row (X=0) with 1s.
* **XY'**: This term is true when X=1 and Y=0. So, we fill in the cell where X=1 and Y=0 with a 1.
```
Y
0 1
X 0 1 1
1 1 0
```
3. **Group the 1s:** We want to make the largest possible groups of 1s, where the groups are powers of 2 (1, 2, 4).
* **Group 1:** The two 1s in the top row form a group of 2. This group represents X' (because X=0 for both cells).
* **Group 2:** The 1 at X=1 and Y=0 is already as large a group as it can be. This represents XY'.
4. **Write the simplified expression:**
The simplified expression is simply the sum of the groups:
F(X, Y) = X' + XY'
In this particular case, the function is *already simplified*. The K-map just confirms that there are no further simplifications possible.

Question 1192174: The intensity level of sound is measured on a logarithmic scale. The intensity level, β, of sound is defined in the following equation β=10log I/Io where β is measured in decibels, and Io is the intensity of a reference level. The reference level usually taken is the "threshold of hearing", which is 1.0x10^-12 W/m^2.
a) What is the intensity level of the threshold of hearing?
b) What is the intensity level of a whisper if the intensity is 1.0x10^-10 W/m^2?
c) How much louder does a siren at 30 m away, with an intensity of 1.0x10^-2 W/m^2, sound compared to a whisper?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
**Understanding Sound Intensity and Decibels**
Sound intensity is measured on a logarithmic scale using decibels (dB). This scale is based on the equation:
β = 10 log (I / I₀)
where:
* β is the sound intensity level in decibels (dB)
* I is the sound intensity in watts per square meter (W/m²)
* I₀ is the reference intensity, usually the threshold of hearing (1.0 x 10⁻¹² W/m²)
**Calculations**
**a) Threshold of Hearing**
The threshold of hearing is our reference point (I₀ = 1.0 x 10⁻¹² W/m²). Plugging this into the equation:
β = 10 log (1.0 x 10⁻¹² / 1.0 x 10⁻¹²) = 10 log (1) = 0 dB
**The intensity level of the threshold of hearing is 0 dB.**
**b) Whisper**
A whisper has an intensity of 1.0 x 10⁻¹⁰ W/m². Calculating the decibel level:
β = 10 log (1.0 x 10⁻¹⁰ / 1.0 x 10⁻¹²) = 10 log (100) = 20 dB
**The intensity level of a whisper is 20 dB.**
**c) Siren**
A siren at 30 meters has an intensity of 1.0 x 10⁻² W/m². Calculating the decibel level:
β = 10 log (1.0 x 10⁻² / 1.0 x 10⁻¹²) = 10 log (10¹⁰) = 100 dB
**The intensity level of the siren is 100 dB.**
**Comparison**
To find how much louder the siren is compared to the whisper, we subtract their decibel levels:
Difference = 100 dB - 20 dB = 80 dB
**The siren sounds 80 dB louder than the whisper.**

Question 1209426: If log base a =p,log base a=q ,express log base a (6) in term of p and q

Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39616)   (Show Source):
You can put this solution on YOUR website!
The exercise is missing the logarithm of WHAT, to the base a.
A guess is you have this:
and .

Then you want to find what is .

and substituting using what is "given",
.

Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!

Re-post, typing the whole question using your keyboard.

Whatever you did in making this post, required information did not show up.

log base a of ?? is p?
log base a of ?? is q?

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.

Good morning.

your post is incomplete and, THEREFORE, is DEFECTIVE.

Please do not submit defective posts to this forum.

Thank you.

Question 1209355: Find x if \log_2 (\log_3 x) = \log_4 x.
Answer by yurtman(42)   (Show Source):
You can put this solution on YOUR website!
**1. Change Bases**
* **Use the change-of-base formula:**
* log_a(b) = log_c(b) / log_c(a)
* Apply this to both sides of the equation:
* log₂(log₃(x)) = log₄(x)
* log₂(log₃(x)) = log₂(x) / log₂(4)
* log₂(log₃(x)) = log₂(x) / 2
**2. Simplify**
* Multiply both sides by 2:
* 2 * log₂(log₃(x)) = log₂(x)
* Apply the power rule of logarithms:
* log₂((log₃(x))²) = log₂(x)
**3. Equate Arguments**
* Since the bases of the logarithms are the same (base 2), we can equate the arguments:
* (log₃(x))² = x
**4. Solve for x**
* This equation is difficult to solve algebraically.
* **Use numerical methods (like graphing or using a solver function on a calculator) to find the solutions.**
* **Solutions:**
* x ≈ 1
* x ≈ 8.51
**Therefore, the possible values of x that satisfy the equation log₂(log₃(x)) = log₄(x) are approximately 1 and 8.51.**

Question 1209354: Suppose a\neq 0. Compute log_{8a} 4b if a = 1 and b = 32.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Change of base rule.

Plug in a = 1 and b = 32.

Rewrite 128 and 8 as powers of 2.

Use the rule log(A^B) = B*log(A)

--------------------------------------------------------------------------

Another approach

Since the bases of the previous equation are equal, the exponents must be equal.

--------------------------------------------------------------------------

Answer: 7/3

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.

To answer this question, I should know what this writing means log_{8a} 4b but from the post and from your non-mathematical writing I can not get it. These dances with hieroglyphs is not Math.

Question 1209313: find log5^6.4
Found 2 solutions by math_tutor2020, Edwin McCravy:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

If you are evaluating the log base 5 of 6.4, then
due to the change of base rule
You can use logs of any valid base with the change of base rule.

This means you could say

where Ln refers to the natural log

Or if you are trying to compute log(5^6.4) then use a calculator to get roughly 4.473408
This assumes the log is base 10.

Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
4.473409028

Question 1209111: 6250/41.02*0.207*0.064√
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
6250/41.02*0.207*0.064√
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

What is written here ?

Question 1208985: Express in index notation, the log of is 64 to base 8 is 2
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
Express in index notation, the log of 64 to base 8 is 2
~~~~~~~~~~~~~~~~~

The correct and smooth Math formulation is THIS :

Express in index notation = 2.

ANSWER.      = 64.

Solved.

Question 1208782: If log(2×+1)-log(3×-2)=1 find ×
Answer by josgarithmetic(39616)   (Show Source):
You can put this solution on YOUR website!
Assumed, that symbol really should be variable, x.

You want base 10?

but you should check if this makes sense in your original equation.

Question 1208746: Log5 (2x)=3
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.

= 3 is the same as 2x = = 125, which implies x = = 62.5. ANSWER

Solved.

-------------------

On logarithms and their properties, see these introductory lessons
    - WHAT IS the logarithm
    - Properties of the logarithm
    - Change of Base Formula for logarithms
    - Evaluate logarithms without using a calculator
    - Simplifying expressions with logarithms
    - Solving logarithmic equations
in this site.

Learn the subject from there.

Question 1208680: Given that log 27+log base b 4=5 and log base a 27-log base a 4=1 without using tables find the values of a and b

Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!

When you want help with a problem, take the time to make sure you are making your post correctly.

Given that log 27+log base b 4=5 and log base a 27-log base 4=1

Add the two equations:

Substitute a=3 in either original equation:

ANSWERS: a=3, b=2

Question 1208672: Given that log4=0.6020 and log3=0.771, use the laws of logathrimns to find the log0.75 and log81

Found 4 solutions by greenestamps, ikleyn, Edwin McCravy, Alan3354:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!

Pay attention to what you are posting before you post it.

log(3) is 0.4771, not 0.771.

Then, using basic rules for logarithms....

log(0.75) = log(3/4) = log(3) - log(4)

log(81) = log(3^4) = 4*log(3)

You can do the arithmetic to get the answers.

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
Given that log4=0.6020 and log3=0.771, use the laws of logathrimns to find the log0.75 and log81
~~~~~~~~~~~~~~~~~~~~~~~~

The problem's formulation in your post is FATALLY WRONG.

log(3) is definitely less than log(4), but your post, by mistake,
provides the value of log(3) GREATER than log(4).

Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!

log(0.75) = log(75/100) = log(3/4) = log(3)-log(4) = you finish log(81) = log(3⁴) = 4log(3) = you finish Edwin

Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Given that log4=0.6020 and log3=0.771, use the laws of logathrimns to find the log0.75 and log81
---------------------
log(3) is not 0.771

Question 1208655: Given that 10^(2x)=0.2 and log5=0.6990, find value of x

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
Given that 10^(2x)=0.2 and log5=0.6990, find value of x
~~~~~~~~~~~~~~~~~~

Notice that 0.2 = 1/5. Therefore, log(0.2) = = log(1) - log(5) = 0 - 0.6990 = -0.6990. Now, we are given = 0.2. Take logarithm base 10 of both sides. You will get 2x = log(0.2) = as we deduced above = -0.699. Hence, x = -0.6990/2 = -0.3495. ANSWER CHECK. = 0.2000 (rounded), which confirms the solution.

Solved.

Question 1208397: Solve for x
log(base 25)(x+1) + log(base 25)(x-3) = 1/2
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
Solve for x
log(base 25)(x+1) + log(base 25)(x-3) = 1/2
~~~~~~~~~~~~~~~~~~~~~

                        Step by step

+ = = Use the fact that = and replace in the right side of the previous equation by = . You will get = . It implies (x+1)*(x-3) = 5 x^2 + x - 3x - 3 = 5 x^2 - 2x - 8 = 0 (x^2 - 2x + 1) - 8 = 1 x^2 - 2x + 1 = 1 + 8 (x-1)^2 = 9 x-1 = +/- = +/- 3. Hence, x is either -2 or 4. Negative value x= -2 does not work in the original equation. Hence, x= 4 is the only solution. ANSWER. The only solution is x= 4.

Solved.

Question 1208368: log16/log32
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
log16/log32
~~~~~~~~~~~~~~~~~~~~~~~~

= = = reduce log(2) = = 0.8. Interesting that it is true for any base of logarithms.

Solved.