Questions on Algebra: Logarithm answered by real tutors!

Tutors Answer Your Questions about logarithm (FREE)

Question 173580: Log6x+log6(x-5)=2
Check for inadmissible roots

Answer by MathTherapy(10858)

You can put this solution on YOUR website!

Log6x+log6(x-5)=2
Check for inadmissible roots
****************************
Respondent, nycsub_teacher(90) provides the following solution,
"Log6x+log6(x-5)=2
We write it like this:
log_6(x/(x-5) = 2
x/(x-5) = 6^2
x/(x-5) = 36," and then asks, "Can you finish now?"

Whether or not the person can finish doesn't really matter, because the solution that'll be derived from the respondent's attempt, will be WRONG!!

Correct way: 
The smaller of the 2 log-variable arguments, x - 5 MUST be > 0, and so, x - 5 > 0____x > 5
So, we get: , with x > 5
                                  ---- Applying  = 
                                             -- Converting from LOGARITHMIC form to EXPONENTIAL form 
                                           
                                    
                                 (x - 9)(x + 4) = 0 ---- FACTORIZING the trinomial
                                  x - 9 = 0      or     x + 4 = 0 ---- Setting each factor equal to 0 (zero)
                                        x = 9      or            x = - 4 (IGNORE)

As one of the 2 values, x = - 4, is NOT > 5, it is an INADMISSIBLE root, or an EXTRANEOUS solution.

Therefore, the SOLE ADMISSIBLE/VALID/ACCEPTABLE solution to this equation is x = 9.

Question 178934: simplify
ln(x^3-x^2)
Answer by ikleyn(53937)

(Show Source):

You can put this solution on YOUR website!
.
simplify ln(x^3-x^2).
~~~~~~~~~~~~~~~~~~~~~~~~~

The formula deriving and the answer in the post by @HyperBrain both are incorrect.

The correct solution is THIS

ln(x^3-x^2) = ln(x^2*(x-1)) = 2ln(x) + ln(x-1) for x > 1. ANSWER

Question 686655: log(x-44)-log7=log(x-14)-logx solve.
also this question logx+log(x-3)=1
Answer by MathTherapy(10858)

(Show Source):

You can put this solution on YOUR website!

log(x-44)-log7=log(x-14)-logx solve.

also this question logx+log(x-3)=1
**********************************
This respondent, @MathLover1(20855) did the same thing that @HyperBrain(694) did  for Problem, logarithm/105757. She used the QUADRATIC
EQUATION formula to solve both quadratic equations, which is UNNECESSARY, since the resultant TRINOMIAL is FACTORABLE! Furthermore, the
problem, log(x - 44) - log 7 = log (x - 14) - log (x), has x - 44 as its SMALLEST log variable-argument. This MUST BE > 0, which means that x > 44. 

So, the equation, log(x - 44) - log 7 = log (x - 14) - log (x), will have the following criteria, x > 44.

She got 2 values for x: x = 49, and x = 2. Obviously, 2 is NOT > 44, so x = 2 is an EXTRANEOUS solution, which makes x = 49, VALID/ACCEPTABLE,
and the SOLE SOLUTION.

Question 105757: Solve logx + log(x-48) = 2
Answer by MathTherapy(10858)

(Show Source):

You can put this solution on YOUR website!

Solve logx + log(x-48) = 2
**************************
As usual, HyperBrain(694)'s solution is PARTIALLY-CORRECT/WRONG.
Furthermore, the TRINOMIAL,  can be factorized into (x - 50)(x + 2), so, using the QUADRATIC EQUATION formula is UNNECESSARY,
as it NOTICEABLY generates LARGE numbers, which most of us deplore!! Unless, of course, one wishes to do so! 

log (x) + log (x - 48) = 2
Finally, as the SMALLER of the 2 log variable-arguments, x - 48 MUST be > 0, we get: x - 48 > 0_____x > 48.
So, equation becomes: log (x) + log (x - 48) = 2, with x > 48.

One of the respondent's 2 solutions, - 2 is NOT > 48, which makes it, EXTRANEOUS. Obviously, the other solution, x = 50 is EFINITELY > 48, making
it VALID/ACCEPTABLE, and therefore, the SOLE SOLUTION.

Question 207535: Solve for x where x is a real number
log x + log (x-3) = 1
Answer by ikleyn(53937)

(Show Source):

You can put this solution on YOUR website!
.
Solve for x where x is a real number
log x + log (x-3) = 1
~~~~~~~~~~~~~~~~~

In his post, @HyperBrain derives two solutions x = 5 and/or x = -2.

But the negative solution can not be accepted: it must be rejected,
since logarithm does not tolerate negative arguments !

Question 613455: How do you find log(x+21)+logx=2?
Answer by MathTherapy(10858)

(Show Source):

You can put this solution on YOUR website!

How do you find log(x+21)+logx=2?
*********************************
Respondent @radh(108) went along a very long path, and arrived at a PARTIALLY-CORRECT solution set. 

The smaller of the 2 logarithmic variable-arguments, x, MUST be > 0, so x > 0
We then get: log (x + 21) + log (x) = 2, with x > 0
  log x(x + 21) = 2 ---- Applying  = 

           ---- Converting to EXPONENTIAL form
          
 
(x - 4)(x + 25) = 0 --- Factoring above TRINOMIAL
  x - 4 = 0     or       x + 25 = 0
        x = 4     or                x = - 25 (IGNORE)

The value, 4, for x, is > 0, but - 25 is NOT! Therefore, ONLY x = 4 is valid/ACCEPTABLE, while - 25 is an EXTRANEOUS solution.

Question 105760: Solve log(x-1)- log6 = log(x-2)- logx
Answer by MathTherapy(10858)

(Show Source):

You can put this solution on YOUR website!

Solve log(x-1)- log6 = log(x-2)- logx
*************************************
@HyperBrain(694)'s answer, below, is WRONG, and DOUBLE-WRONG for the negative value of x, since x MUST BE > 2.
" or "
****************
The smallest log variable-argument, x - 2 MUST BE > 0, so x > 2. We then get:
, with x > 2.
           ----- Applying  = 
                      ----- Applying e = f, since  = 
                  x(x - 1) = 6(x - 2) --- Cross-multiplying
                     
      
            
          (x - 4)(x - 3) = 0
           x - 4 = 0       or       x - 3 = 0 ---- Setting each factor equal to 0
                 x = 4       or            x = 3

Both values of x are VALID/ACCEPTABLE, since both are > 2.

Question 742189: Solve and check
a) 2log(x-1) = 2+ log 100
Answer by MathTherapy(10858)

(Show Source):

You can put this solution on YOUR website!

Solve and check
a) 2log(x-1) = 2+ log 100
*************************
The response from @josgarithmetic(39831), while correct, is confusing, in this author's opinion. Whenever one sees a log argument without a base,
it's widely known that that base is 10, since that's the base/number system we all work in. This is just the same as x. It's known that this is actually
1x, but it's not written like that, just x. Also, x is actually , but again, it's not written as such, just x. So, why confuse a person by working the
problem without a base, just because a base wasn't stated? That base, as stated before, is OBVIOUSLY no other than 10. 

          2 log(x - 1) = 2 + log (100)
          2 log(x - 1) = 2 + 2 ---- log (100) = 2
          2 log(x - 1) = 4
 ----- Dividing each side by 2 
            log (x - 1) = 2
                      ---- Converting to EXPONENTIAL form
                     x - 1 = 100
                           x = 100 + 1 = 101

You can do the CHECK!!

Question 946643: Solve fora in log(2a)- 3log2= 1/2log(a-3)
Found 2 solutions by MathTherapy, ikleyn:
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Solve fora in log(2a)- 3log2= 1/2log(a-3) ***************************************** The SMALLEST log argument of the 3 log arguments, log (a - 3), signifies that a - 3 > 0, and so, a MUST BE > 3. , with a being > 3. 2 log (2a) - 6 log (2) = log (a - 3) ----- Multiplying by LCD, 2 2 log (2a) - log (a - 3) = 6 log (2) --- Subtracting log (a - 3) and adding 6 log (2), to both sides ---- Applying = ----- Applying = ---- If , then c = d ---- Cross-multiplying ---- Dividing each side by 4 (a - 12)(a - 4) = 0 --- Factorizing TRINOMIAL a = 12 > 3, and a = 4 > 3. Therefore, 12 and 4 are VALID/ACCEPTABLE values of a.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Solve for a in log(2a)- 3log2= 1/2log(a-3).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        @lwsshar3 in his post gives the answer "no real solutions".
        This answer is incorrect.
        I came to bring a correct solution.

log(2a)- 3log2 = 1/2*log(a-3)
log(2a/(2^3)) = log((a-3)^(1/2))
2a/8 = (a-3)^(1/2)
a/4 = sqrt(a-3)
a = 4*sqrt(a-3)
square both sides:
a^2 = 16*(a-3)
a^2 = 16a - 48
a^2 - 16a + 48 = 0
Discriminant d = b^2-4ac = (-16)^2 - 4*16*48 = 64

= =         (quadratic formula)

The solutions are a = 12 and a = 4.        ANSWER

Both solutions satisfy the original equation.

Solved correctly.

Question 946654: If and , express in terms of x and y.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
If and , express in terms of x and y.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The answer 'x' in the post by @lwsshar3 is INCORRECT.

The correct answer is '-x'.

Question 957757: Suppose log a = 0.3, log b = -0.5. Evaluate each of the following. Don't just give a numerical
answer; indicate clearly the log rules you used to calculate the answer.
i.
ii.

Thank you
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!

Question 957758: Find the domain of the given function
f(x)=(√lnx)
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Find the domain of the given function
f(x)=(√lnx)
~~~~~~~~~~~~~~~~~~~~~~~~~~~

In his post, @lwsshar3 makes a note

        for x < 1, ln(x) < 0 (this is true for logs of any base)

As an absolute statement, this note is wrong: for x < 1, is negative only if the base 'a' is greater than 1 (one).

In opposite, if the base 'a' is lesser than 1, then is POSITIVE for x < 1.

Be aware.

Question 970399: prove that log 2 base 10 greater that 0.3
Answer by n2(91)   (Show Source):
You can put this solution on YOUR website!
.
prove that log 2 base 10 greater than 0.3.
~~~~~~~~~~~~~~~~~~~~~~~~~~

            There is a proof of striking beauty.

Start from this inequality 1024 > 1000, which obviously is valid.

Rewrite it in the form      > .

Take logarithm base 10 of both sides:     10*log(2) > 3.

Divide both sides by   10:     log(2) > ,   or     log(2) > 0.3.

Thus the statement is proven.

Question 481624: Solve:
2 * log--base4 (x) + 3 * log--base8 (x)=10
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!

Use on both log expressions.

Use on both log expressions.

ANSWER: x = 32

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Solve: 2 * log--base4 (x) + 3 * log--base8 (x)=10 ****************************************** Respondent @wsshak3(11628) came up with a decimal value for x, but this problem clearly has an INTEGER-solution, as seen below!! While both are close to each other, the approximated decimal-answer doesn't QUITE MAKE the equation true, but ALMOST TRUE, while the INTEGER-value for x, below, does. ---- Converting ALL logs to base 2 ----- Converting to EXPONENTIAL form

Question 1149204: If interest is compounded continuously at the rate of 3% per year, approximate the number of years it will take an initial deposit of $7000 to grow to $27,000. (Round your answer to one decimal place.)

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
If interest is compounded continuously at the rate of 3% per year, approximate the number of years
it will take an initial deposit of $7000 to grow to $27,000. (Round your answer to one decimal place.)
~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @addingup to both problems are incorrect.
        It is because his starting setup equation is incorrect, which is a conceptual/strategic error.

        I came to bring a correct solution.

The setup equation for this continuously compounded account is 27000 = , <<<---=== It is a standard translation for a continuously compound account where 't' is the time in years. Divide both sides by 7000 to get = . Take natural logarithm of both sides = 0.03*t. Express 't' and calculate t = = 44.9976 years. It is reasonable to round it to the closest integer, which is 45 years. ANSWER. 45 years.

Solved correctly, so you can learn/teach safely from my post.

Ignore the post by @addingup, since it is wrong solution and wrong teaching.

Question 1163696: It is believed that two quantities, z and d are Connected by the relationship of the form z=kd^n where k and n are provided that d doesn't exceed some fixed (but unknown)values D.An experiment produced the following data
D 750 810 870 930 990 1050 1110 1170
Z 2.1 2.6 3.2 4.0 4.8 5.6 5.9 6.1
a)Plot the values of log10Z against log10 d.Use these points to suggest a
value for D.
b)It is known tht for d < D,n is a whole number.Use your graph to find the value of n.Show also that k=5×10^-9
c) Use your value of n and the estimate k=5×10^-9 to find the value of d for which z=30.
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import numpy as np
import matplotlib.pyplot as plt
d = np.array([750, 810, 870, 930, 990, 1050, 1110, 1170])
z = np.array([2.1, 2.6, 3.2, 4.0, 4.8, 5.6, 5.9, 6.1])
log_d = np.log10(d)
log_z = np.log10(z)
plt.scatter(log_d, log_z, color='blue', label='Data points')
plt.plot(log_d, log_z, 'b--', alpha=0.3)
plt.xlabel('log10(d)')
plt.ylabel('log10(z)')
plt.title('log10(z) vs log10(d)')
plt.grid(True)
plt.savefig('log_plot.png')
# Calculate slopes between consecutive points
slopes = np.diff(log_z) / np.diff(log_d)
print(f"Slopes: {slopes}")
# Let's perform linear regression for the first few points to check n
# Trying first 5 points (d = 750 to 990)
for i in range(3, 8):
subset_log_d = log_d[:i]
subset_log_z = log_z[:i]
m, c = np.polyfit(subset_log_d, subset_log_z, 1)
print(f"Points 1 to {i}: slope (n) = {m:.4f}, intercept (log10 k) = {c:.4f}, k = {10**c:.2e}")

```
```text?code_stdout&code_event_index=2
Slopes: [2.7750937 2.90571474 3.34591322 2.91619507 2.61980574 0.93910191
0.63324575]
Points 1 to 3: slope (n) = 2.8372, intercept (log10 k) = -7.8355, k = 1.46e-08
Points 1 to 4: slope (n) = 2.9825, intercept (log10 k) = -8.2565, k = 5.54e-09
Points 1 to 5: slope (n) = 3.0028, intercept (log10 k) = -8.3155, k = 4.84e-09
Points 1 to 6: slope (n) = 2.9634, intercept (log10 k) = -8.2005, k = 6.30e-09
Points 1 to 7: slope (n) = 2.7732, intercept (log10 k) = -7.6442, k = 2.27e-08
```
[image-tag: code-generated-image-0-1776447487869153342]
To find the values of $k$, $n$, and $D$ for the relationship $z = k d^n$, we transform the equation by taking the logarithm (base 10) of both sides:
$$\log_{10} z = n \log_{10} d + \log_{10} k$$
This is a linear equation of the form $y = mx + c$, where the gradient is $n$ and the y-intercept is $\log_{10} k$.
### a) Plotting and suggesting a value for $D$
First, we calculate the values for $\log_{10} d$ and $\log_{10} z$:
| $d$ | $750$ | $810$ | $870$ | $930$ | $990$ | $1050$ | $1110$ | $1170$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $z$ | $2.1$ | $2.6$ | $3.2$ | $4.0$ | $4.8$ | $5.6$ | $5.9$ | $6.1$ |
| $\log_{10} d$ | $2.875$ | $2.908$ | $2.939$ | $2.968$ | $2.996$ | $3.021$ | $3.045$ | $3.068$ |
| $\log_{10} z$ | $0.322$ | $0.415$ | $0.505$ | $0.602$ | $0.681$ | $0.748$ | $0.771$ | $0.785$ |
**Analysis of the Plot:**
When plotting these points, we look for where the linear trend stops. Calculating the gradient (slope) between successive points shows:
* From $d=750$ to $d=1050$, the slope remains consistently near **$3.0$**.
* After $d=1050$, the slope drops drastically to $\approx 0.96$ (between 1050-1110) and $\approx 0.63$ (between 1110-1170).
**Suggested Value for $D$:**
Based on the graph, the linear relationship holds up to **$D \approx 1050$**. Beyond this value, the data points clearly deviate from the straight line.
---
### b) Finding $n$ and showing $k = 5 \times 10^{-9}$
**Determining $n$:**
Since $n$ is known to be a whole number for $d < D$, we look at the slopes of the linear portion:
* Slope ($870$ to $930$) $\approx \frac{0.602 - 0.505}{2.968 - 2.939} = \frac{0.097}{0.029} \approx 3.3$
* Slope ($930$ to $990$) $\approx \frac{0.681 - 0.602}{2.996 - 2.968} = \frac{0.079}{0.028} \approx 2.8$
The average slope is approximately 3. Thus, **$n = 3$**.
**Showing $k = 5 \times 10^{-9}$:**
Using the formula $k = \frac{z}{d^3}$ for values where $d < D$:
* For $d=750$: $k = \frac{2.1}{750^3} \approx 4.98 \times 10^{-9}$
* For $d=930$: $k = \frac{4.0}{930^3} \approx 4.97 \times 10^{-9}$
* For $d=990$: $k = \frac{4.8}{990^3} \approx 4.95 \times 10^{-9}$
Rounding to one significant figure, we find **$k = 5 \times 10^{-9}$**.
---
### c) Find the value of $d$ for which $z=30$
Using the model $z = (5 \times 10^{-9})d^3$, we substitute $z = 30$:
$$30 = (5 \times 10^{-9}) d^3$$
$$d^3 = \frac{30}{5 \times 10^{-9}}$$
$$d^3 = 6 \times 10^9$$
$$d = \sqrt[3]{6 \times 10^9}$$
$$d = 1000 \times \sqrt[3]{6}$$
$$d \approx 1000 \times 1.817$$
**$d \approx 1817$**
*(Note: While $1817 > D$, the calculation based strictly on the suggested constants yields this result.)*

Question 289032: Can you please help me with this question (the steps to finding the answer)
Solve algebraically for t:
4^(-3t) = .10
Thank you!
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Can you please help me with this question (the steps to finding the answer)
Solve algebraically for t:
4^(-3t) = .10
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The derivation of the solution in the post by @mananth is incorrect.
        I came to do this job as accurately as it should be done.

Your starting equation is = 0.1. Take logarithm base 10 of both sides (-3t)*log(4) = log(0.1), (-3t)*log(4) = -1, t = = = 0.55365 (approximately). ANSWER

Solved correctly to teach you in a right way.

Question 686407: Please help me solve this equation:
Solve: log2 (x+5) + log2 (x+1) = 5
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Please help me solve this equation: Solve: log2 (x+5) + log2 (x+1) = 5 *********************************** The following response from the other person doesn't do much to answer this problem, or help the "POSTER: "log(x+1)/log 2 +log*x+5)/log(2)=5" , with x > - 1 ------ Applying = ---- Converting to EXPONENTIAL form (x - 3)(x + 9) = 0 x - 3 = 0 or x + 9 = 0 ---- Setting each FACTOR equal to 0 x = 3 or x = - 9 (IGNORE) The above x-value, - 9, is IGNORED because x MUST be > - 1, and x = - 9 is CLEARLY NOT! So, only VALID/ACCEPTABLE solution is: x = 3

Question 287532: Solve Log2(x+1)=5
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Solve Log2(x+1)=5 ***************** The other person who responded has the following, which makes ABSOLUTELY no SENSE, in this author's opinion! "log ( 2x + 2 ) = 5 => e^5 = 2x + 2 => 2x = e^5 -2 Now do the rest.... Happy" , with x > - 1 ------ Converting to EXPONENTIAL form x + 1 = 32 x = 32 - 1 = 31 That's IT!!

Question 865839: - + = 2
Please solve for x
Please show all steps
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

- + = 2 Please solve for x Please show all steps ********************* While the answer is correct, the steps the other person showed are WAY TOO UNNECESSARY. For example, why did he convert from base 4 to base 2? To this author, it's a little lengthy, and quite time-consuming. A lot less steps can be used to reach the same goal: the correct answer! ----- Applying ----- Applying = ---- Converting to EXPONENTIAL form 16x = y ----- Cross-multiplying

Question 734457: log10x-log(x-3)=log2
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log10x-log(x-3)=log2 ******************** is NOT a solution to this logarithmic equation, as stated by the other person, since it's EXTRANEOUS! log (10x) - log (x - 3) = log (2) x - 3 is the smaller of the 2 variable-arguments, and so, MUST be > 0. This gives us: x - 3 > 0 ====> x > 3 We then have: log (10x) - log (x - 3) = log (2), with x > 3 While the solution to this equation is x = , this value is NOT > 3, and so, is INVALID/UNACCEPTABLE. As a result, NO SOLUTION exists for this equation!.

Question 883161: log2(x - 6) + log2(x - 4) = log2 x solve the logarithmic equation. and round.
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log2(x - 6) + log2(x - 4) = log2 x solve the logarithmic equation. and round. **************************************************************************** x = 3 or x = 8, as stated by the other person, is partly WRONG! Looking at the logarithmic equation, the smallest log argument, x - 6 MUST be > 0. So, x - 6 > 0, and therefore, x > 6. We then have: , with x > 6 The x-value 3 is NOT > 6, which makes it EXTRANEOUS and an INVALID/UNACCEPTABLE solution. On the other hand, the x-value 8 is > 6, which makes it the ONLY solution to this equation.

Question 954792: Solve log2(x-2)-log2(x+5)=3
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Solve log2(x-2)-log2(x+5)=3 ***** The solution, x = - 6, by the other person who responded, is WRONG!! The SMALLER of the 2 log arguments, x - 2 MUST be > 0. So, x - 2 > 0 ===> x > 2. We then have: , with x > 2. ----- Applying = ---- Converting to EXPONENTIAL form 8(x + 5) = x - 2 ----- Cross-multiplying 8x + 40 = x - 2 8x - x = - 2 - 40 7x = - 42 The x-value, - 6, is NOT > 2, and is therefore an EXTRANEOUS solution, which makes it an INVALID/UNACCEPTABLE solution. So, this equation DOESN'T have a solution!!

Question 22137: log2x+log2(x-6)=4
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log2x+log2(x-6)=4 ***************** The solution, (x = 8, or x = - 2) by the other person who responded, is PARTIALLY WRONG!! The SMALLER log argument, x - 6, MUST be > 0. So, x - 6 > 0 ===> x > 6. We then have: , with x > 6. ----- Applying = --- Converting to EXPONENTIAL form <=== Note that the other person has i/o , but both have the same value, 16 (x - 8)(x + 2) = 0 x - 8 = 0 OR x + 2 = 0 ---- Setting each FACTOR equal to 0 x = 8 OR x = - 2 The x-value, 8, is > 6, but - 2 is NOT. This makes - 2 an EXTRANEOUS solution!! So, x = 8 is the only VALID/ACCEPTABLE solution!!

Question 735204: Please show how to solve:
Log base10(n^2 � 90n) = 3
Thank you
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Please show how to solve: Log base10(n^2 � 90n) = 3 Thank you ************************************* To this author, what the other person who responded wrote, as a solution to this equation, doesn't make any sense. That's, "....n^2 - 90n = log base 10 (3) = .48 use quadratic formula and we get (note that b is --90 or 90) n=(90+square root (90^2-4*1*-.48)) / 2 = 90" <== Base 10 is OPTIONAL, because we work in the decimal system, so usually, base 10 is NOT entered. Since the log argument MUST be greater than 0, we have: ===> . The SOLUTIONS to the INEQUALITY, 0 and 90 are the CRITICAL POINTS, with 3 intervals: Interval 1: n-values < 0 Interval 2: 0 < n-values < 90 Interval 3: n-values > 90 When tested, we find that the n-values that'll satisfy the INEQUALITY are < 0, and > 90. So, based on that, we get: , with n < 0, or > 90. ---- Converting to EXPONENTIAL form (n - 100)(n + 10) = 0 n - 100 = 0 OR n + 10 = 0 ----- Setting each factor equal to 0 n = 100 OR n = - 10 As seen, 100 is > 90, and - 10 is < 0, so both solutions are VALID/ACCEPTABLE!

Question 1111913: Rachel Spender wants to invest $4000 in savings certificates which bear an interest rate of 7.25% compounded semi-anually. How long a time period should she choose in order to save an amount of $4700?
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Rachel Spender wants to invest $4000 in savings certificates which bear an interest rate of 7.25% compounded semi-anually. How long a time period should she choose in order to save an amount of $4700? ***************************************************** Future-value-of-$1 formula: , with = Future Value (Unknown, in this case) = Principal/Initial Deposit ($4,000, in this case) = Interest rate, as a decimal (7.25%, or .0725, in this case) = Number of ANNUAL compounding periods (semiannually, or 2, in this case) = Time Principal/Initial Deposit has been invested, in YEARS (t, in this case) How long a time period should she choose in order to save an amount of $4700? ----- Substituting $4,700 for A, $4,000 for P, .0725 for i, and 2 for m ----- Converting to LOGARITHMIC form Time it'll take the $4,000 investment to increase to $4,700, or = 2.26446601 years, which needs to be ROUNDED UP to years, or 2 years, 6 months. ** Notice that although 2.26446601 rounds off to about years, the $4,000 investment, at the -year juncture, will increase to about $4,695.16 (< $4,700). This is why it's necessary to ROUND UP to year , or 2.5 years (at the semi-annual point), at which time, the $4,000 initial deposit will exceed $4,700 (about $4,779.50, to be exact).

Question 1112016: A roasted turkey is taken from an oven when its temperature has reached 185 Fahrenheit and is placed on a table in a room where the temperature is 75 Fahrenheit.
a). If the temperature of the turkey is 141 Fahrenheit after half an hour, what is its temperature after 45 minutes?
b). When will the turkey cool to 100 Fahrenheit?
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

A roasted turkey is taken from an oven when its temperature has reached 185 Fahrenheit and is placed on a table in a room where the temperature is 75 Fahrenheit. a). If the temperature of the turkey is 141 Fahrenheit after half an hour, what is its temperature after 45 minutes? b). When will the turkey cool to 100 Fahrenheit? ************************************************ You can use Newton's Law of Cooling.......any of the following 3 formulae should work: This author's preference is the 1^st formula. In this case, the cooling rate (k) is first needed, and is derived as follows: , where: = time taken to get to a COOLED temperature ( an hr, or 30 minutes, in this case) = TEMPERATURE (T) at a given time (t)___(141^oF, in this case) = SURROUNDING Temperature (75^oF, in this case) = ORIGINAL/INITIAL temperature (185^oF, in this case) = the CONSTANT or COOLING rate (UNKNOWN, in this case) ---- Substituting 30 for t, 75^o for , and 185^o for ----- Substituting 141^o for T(30) ----- Converting to NATURAL LOGARITHMIC (ln) form ****************************************************************************************** a). If the temperature of the turkey is 141 Fahrenheit after half an hour, what is its temperature after 45 minutes? ---- Substituting 45 for t, 75^o for , 185^o for , and .01702752 for k Temperature, after 45 minutes, or , or approximately 126^oF. ****************************************************************************************** b). When will the turkey cool to 100 Fahrenheit? ---- Substituting 100 for T(t), 75^o for , 185^o for , and .01702752 for k ---- Converting to NATURAL LOGARITHMIC (ln) form Time taken for the turkey to cool to 100^oF, or

Question 1130119: Could someone please assist with the question:
A pot of boiling soup with an internal temperature of 100� Fahrenheit was taken off the stove to cool in a 69�F room. After fifteen minutes, the internal temperature of the soup was 93�F.
To the nearest minute, how long will it take the soup to cool to 81�F?
Here is my work:
69+(100-69) * e= 69+31*e
93=69+31*e(-k*15)
31*e^(-k *15)= 93-69= 24
e^-15k= 24/31= .7742
-15K= ln(.7742)
k= -ln(0.7742)/ 15=0.017
T(t)= 69+31*e (0.017)=81
e(-0.017*t)= 81-69/31=.387
-0.017*t= ln(.387)
t= - ln (.387)/ (0.017)= 55.8 min

Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.

Hello, in your post (in your problem's formulation), there is a fatal error,
which equates the problem's creator to zero.

Concretely, the boiling temperature of 100 degrees Celsius is missed with 100 degrees Fahrenheit,
which has no any relation to boiling.

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Could someone please assist with the question: A pot of boiling soup with an internal temperature of 100� Fahrenheit was taken off the stove to cool in a 69�F room. After fifteen minutes, the internal temperature of the soup was 93�F. To the nearest minute, how long will it take the soup to cool to 81�F? Here is my work: 69+(100-69) * e= 69+31*e 93=69+31*e(-k*15) 31*e^(-k *15)= 93-69= 24 e^-15k= 24/31= .7742 -15K= ln(.7742) k= -ln(0.7742)/ 15=0.017 T(t)= 69+31*e (0.017)=81 e(-0.017*t)= 81-69/31=.387 -0.017*t= ln(.387) t= - ln (.387)/ (0.017)= 55.8 min ********************************* You can use Newton's Law of Cooling.......any of the following 3 formulae should work: This author's preference is the 1^st formula. In this case, the cooling rate is first needed, and is derived as follows: , where: = time taken to get to a COOLED temperature (15 minutes, in this case) = TEMPERATURE (T) at a given time (t)___(93^oF, in this case) = SURROUNDING Temperature (69^oF, in this case) = ORIGINAL/INITIAL temperature (100^oF, in this case) = the CONSTANT or COOLING rate (UNKNOWN, in this case) ----- Substituting 15 for t, 69^o for , and 100^o for ----- Substituting 93^o for T(15) ----- Converting to NATURAL LOGARITHMIC (ln) form *********************************************************************** To the nearest minute, how long will it take the soup to cool to 81�F? ----- Substituting 69^o for , 100^o for , and .0170622 for k ----- Substituting 81^o for T(t) ----- Converting to NATURAL LOGARITHMIC (ln) form Time it takes for the soup to cool to 81^oF, or As your answer, 55.8 min, or approximately 56 mins, coincides with mine, it is correct! Great JOB!!

Question 1163697: unsolved 2020-08-22 10:07:30 It is believed that two quantities, z and d are Connected by the relationship of the form z=kd^n where k and n are provided that d doesn't exceed some fixed (but unknown)values D.An experiment produced the following data
D 750 810 870 930 990 1050 1110 1170
Z 2.1 2.6 3.2 4.0 4.8 5.6 5.9 6.1
a)Plot the values of log10Z against log10 d.Use these points to suggest a
value for D.
b)It is known tht for d < D,n is a whole number.Use your graph to find the value of n.Show also that k=5×10^-9
c) Use your value of n and the estimate k=5×10^-9 to find the value of d for which z=3.0.
Answer by KMST(5396)   (Show Source):
You can put this solution on YOUR website!
If , then
The graph of against would be expected to be a straight line with slope up to a certain value .
We need to calculate and tabulate and

a)Then we plot against
The points in green, up to , corresponding to fit well enough on a line. That supports suggesting .

b) We could estimate as the slope between points and
That slope is
Since is supposed to be a whole number, it must be .

c) For (rounded)
Substituting values for , , with and into we get

to get as the value of d for which z=3.0.

Question 166689: log3(x-2)+log3(x+4)=3
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log3(x-2)+log3(x+4)=3 ********************* The other person who responded is WRONG!! His solutions, "x = {3.243, -5.243}" are WRONG. If he'd checked his answers, he would've realized this. The SMALLER log argument, x - 2 MUST be > 0, so x - 2 > 0, and we get: x > 2. We now have: , with x being > 2. ----- Applying = ---- Converting to EXPONENTIAL form (x - 5)(x + 7) = 0 x - 5 = 0 OR x + 7 = 0 x = 5 OR x = - 7 (ignore) The constraint above "states" that x MUST be > 2. 5 is > 2, but - 7 is NOT. This makes - 7 an EXTRANEOUS solution, and the only ACCEPTABLE solution, x = 5. You can do the CHECK!!

Question 945331:
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Question 1088632: log3 (2x-1) = 2 x =
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log3 (2x-1) = 2 x = ******************** log3 (2x-1) = 2 x = This, most likely is: x =, instead of (what the other person surmised!!). 2x - 1 = 9 2x = 9 + 1 2x = 10 = 5

Question 1026110: simplify the expression
log3 (x+1) -log3 (3x^2-3x-6)+log3 (x-2)
The 3 on each log is lowered
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

simplify the expression log3 (x+1) -log3 (3x^2-3x-6)+log3 (x-2) The 3 on each log is lowered **************************** is NOT the simplified form of log3 (x+1) -log3 (3x^2-3x-6)+log3 (x-2), as the other person who responded, indicates! ----- Applying = ----- Factoring out GCF, 3, in DENOMINATOR ---- Factorizing QUADRATIC/TRINOMIAL, in DENOMINATOR = = = - 1

Question 747802: log3(x+6)+log3(x-6)-log3x=2
solve the logarithmic equation
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log3(x+6)+log3(x-6)-log3x=2 solve the logarithmic equation ****************************** It's heart-wrenching to see how the other person (@MATHLOVER) makes this problem so long and time-consuming! Why?? In my opinion, she should learn how to solve problems much more efficiently. ************************************************************************* This log equation starts with 3 log arguments: a1 (x + 6), a2 (x - 6), and a3 (x), the smallest being log argument a2, or x - 6. Argument "a2" "tells" one that x - 6 MUST be > 0. So, x - 6 > 0_____x > 6 We now have: , with x > 6 ---- Applying = --- Applying , when: --- Cross-multiplying (x - 12)(x + 3) = 0 x - 12 = 0 OR x + 3 = 0 x = 12 OR x = - 3 As stated above, x MUST be > 6, so ONLY x = 12 is ACCEPTABLE. This makes x = - 3, an EXTRANEOUS solution to this equation, and is therefore IGNORED/REJECTED.

Question 815452: if x*x+y*y=7xy, prove
log(x+y)=1/2(logx+logy)+log3
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!

A common "trick" for working on an equation like this, with "" on the left and an expression involving "" on the right, is to add on the left to make a perfect square trinomial.

The problem asks us to find an expression for log(x+y), so take logs of both sides:

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

if x*x+y*y=7xy, prove log(x+y)=1/2(logx+logy)+log3 If x*x+y*y=7xy, prove log(x+y)=1/2(logx+logy)+log3 Prove: ----- Squaring x + y --- Substituting 7xy for (GIVEN) ------- Taking the square root of both sides <=== ONLY the POSITIVE square root on the right-side is needed here ----- Taking the log of both sides of the above equation <=== QED!

Question 769077: if logx =4 and logy =3 Evaluate log square root of x divided by y squared
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!

Algebraic expressions written in words are often ambiguous and open to different interpretations, as the other tutor indicated.

Assuming this is a problem from a student learning basic rules of logarithms, the most likely interpretation of the expression to be evaluated is

In that case, use basic logarithm rules to evaluate the expression.

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

if logx =4 and logy =3 Evaluate log square root of x divided by y squared ========================================================================= Not clear at all!! log (x) = 4___ log (y) = 3___ If it's , then answer is: , or If it's , then answer is: - 4. None of these correspond to the answer from the other person who responded!

Question 717220: log(base x)(1/4) = (-2/3)
log(
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!

Convert to exponential form:

Raise both sides of the equation to the (-3/2) power:

ANSWER: 8

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

log(base x)(1/4) = (-2/3) --------- Converting to EXPONENTIAL form -- RAISing each side to the NEGATIVE reciprocal-exponent on x (to the power)

Question 250018: solve:

Unfortunately I do not know how to put this without all the words. But this is a problem on one of my take home tests. I dont understand how to solve because I get for the log in the parentheses but now I don't knwo what to do?
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

solve: Unfortunately I do not know how to put this without all the words. But this is a problem on one of my take home tests. I dont understand how to solve because I get for the log in the parentheses but now I don't knwo what to do? =============================================== In this problem, x is one of the 3 bases given. So, being a LOG-BASE, x CANNOT be 0, and also, x MUST be GREATER than 1. , with x > 1 --- --- ----- Converting to EXPONENTIAL form However, as stated above, x MUST be > 1, so There are NO SOLUTIONS to this equation. The other person is WRONG, as he/she states that the value of x is 1.

Question 1129970: Find the exact value of e^LOGe^2 16
The answer is 4, but I am unsure as to why.
If possible, please show multiple methods (with step by step breakdown). Here is what I've gathered so far:
Method 1:
1. e^LOGe^2 2^4
2. e^(4/2)LOGe 2 ( I don't understand where the 4 divided by 2 comes into play, and which property dictates it )
3. e^2LOGe 2
4. e^LOGe 4 (Where did the 4 come from? Did both of the 2's get multiplied?)
5. =4
Method 2:
1. e^LOGe^2 16
2. e^(1/2)LN 16 (Where did the 1/2 come from?)
3. sqrt(e^LN 16) (Why is the entire expression inside the square root?)
4. sqrt(16) (Why did the "e^LN" portion inside the square root disappear?)
5. =4
Is it also possible to solve this problem by using the base change method?
Thanks in advance.
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Find the exact value of e^LOGe^2 16 The answer is 4, but I am unsure as to why. If possible, please show multiple methods (with step by step breakdown). Here is what I've gathered so far: Method 1: 1. e^LOGe^2 2^4 2. e^(4/2)LOGe 2 ( I don't understand where the 4 divided by 2 comes into play, and which property dictates it ) 3. e^2LOGe 2 4. e^LOGe 4 (Where did the 4 come from? Did both of the 2's get multiplied?) 5. =4 Method 2: 1. e^LOGe^2 16 2. e^(1/2)LN 16 (Where did the 1/2 come from?) 3. sqrt(e^LN 16) (Why is the entire expression inside the square root?) 4. sqrt(16) (Why did the "e^LN" portion inside the square root disappear?) 5. =4 Is it also possible to solve this problem by using the base change method? Thanks in advance. ================== Find the exact value of e^LOGe^2 16 ===> Let ---- Converting to EXPONENTIAL form ----- Exponents are equal. and so are the bases ---- Back-substituting for t APPLYING , now becomes: , which is SIMPLY 4 OR Find the exact value of e^LOGe^2 16 ===> First of all, let's EXTRACT the exponent, from the expression. APPLYING , now becomes: , which is SIMPLY 4.