Questions on Algebra: Exponent and logarithm as functions of power answered by real tutors!

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Log(base 8) (8^9)^(1/2) = n
Log(base 8) (8^4.5) = n
4.5 * Log(base 8) (8) = n
4.5 = n

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As usual, there are several ways to solve this. Maybe the quickest way is based on

• knowing our exponents (including fractional exponents) well
• seeing that the argument of the base 8 logarithm is a power of 8
• knowing that you do not need a calculator to find the base 8 logarithm of a power of 8

Since square roots are the same as raising to the 1/2 power we can rewrite the argument:

Using the property of exponents, , we get:

Using the property of logarithms, , we can move the exponent out in front:

Since by definition :

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---
g(x)=x^2+3x

h(x)=(x+3)/2

find (h*g)(-1)
============================
Multiply the two algebraic expressions; then evaluate that product at x=-1.
--------------------------------------------
(h*g)(-1) = [(x+3)/2](-1)
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[(x^3+3x^2)/2](-1)
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[(-1)^3+3(-1)^2]/2
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=[-1+3]/2
=[2/2]
= 1
==============
Cheers,
Stan H.
=============================

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If f(x)=e^2x, then
f(1/4) just means to substitute the value of 1/4 in place of the x.
f(1/4)=e^(2*1/4)
f(1/4)=e^(1/2)=e^0.5=1.6487 approximately!

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus

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2,401
1,000,000

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Can you simplify it?( no negative exponent)
(6x^3y^-4z^5)^-2
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= 1/(36x^6y^-8z^10)
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= y^8/(36x^6z^10)
========================

Can you find x?
square root of (64^x) = 256
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(2^6x) = (2^8)
6x = 8
x = 4/3
====================
Cheers,
Stan H.

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Solving equations with variables in the exponents often requires the use of logarithms. Since we are asked to find exact answers, it means that we will not use calculators to find the logarithms. We will just leave the logarithms alone. So it doesn't make any difference which base we use for the logarithms. I'll solve the problem using a couple of different bases:

• Base 4 logarithms.
  • Find base 4 logarithm of each side:
    
  • Use the property to move the exponents in front:
    
  • Since , (This is why I chose base 4 logarithms)
    
  • Now we solve for x. Subtract x from each side (to get the x's on one side):
    
  • Factor out x on the right side:
    
  • Divide both sides by :
    
    which is an exact answer.
• Base 7 logarithms.
  • Find base 7 logarithm of each side:
    
  • Use the property to move the exponents in front:
    
  • Since , (This is why I chose base 7 logarithms)
    
  • Now we solve for x. Use the Distributive property on the left side:
    
  • Subtract x*log(7, (4)) from each side (to get the x's on one side):
    
  • Factor out x on the right side:
    
  • Divide both sides by :
    
    which is an exact answer.
• Other base logarithms. These can be used. But since there will not be a or , the expressions and solution will be even more complicated.

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The keys to these problems are:

• Exponents can be any Real number: zero, positive, negative, fractions, irrational, etc. And since logarithms are exponents they too can be any Real number.
• The result of raising a positive number to a power will always be positive, no matter what the exponent is! (Remember that zero exponents result in 1's and negative exponents do not result in negative values. Negative exponents mean "reciprocal of" and the reciprocal of a positive is also positive.) And since the argument of a logarithm is the result of raising a positive number to a power, the argument must be positive.

Domain of g(x).
Since the x is in the argument of a logarithm, the domain must make the argument of the logarithm positive:

Solving this we get:

which is the domain of g(x).

Range of g(x).
Since g(x) is a logarithm, the range is all Real numbers.

Domain of f(x).
Since x is the exponent, the domain is all Real numbers.

Range of f(x).
Since can be any positive number, from the tiniest fraction (for large negative x's) to the infinitely large (for large positive x's), we just have to figure out what values will be. When is a tiny fraction, f(x) will be a tiny bit less than 1. When is infinitely large, f(x) will be an infinitely large negative number. So the range of f(x) is all Real numbers less than 1.

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See solution to problem 239904

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If the equation is , then read on. If not then repost your question with a correct and clear equation, using parentheses generously.

When a subscript of 0 is used, it is often used to indicate the initial or starting value for that value. In this case, represents the initial amount of the radioactive material. So we can rephrase the problem as: If , after how many years will the amount, A, be 150? So we want to solve the equation:

To solve for variables in the exponent like this, we often use logarithms. But first we will isolate the e and its exponent by dividing both sides by 700:

The fraction on the left will reduce to:

Now we will use logarithms. We can use any logarithm which our calculator can handle. The "natural" choice (excuse the pun) is to use natural logarithms because of the e. But in case your calculator only does base 10 logarithms I will finish the solution using both:

• Natural logarithms.
  • Find the natural logarithm of each side:
    
  • Use the property to move the exponent out in front:
    
  • Since ln(e) = 1:
    
  • Divide both sides by -0.0077:
    
  • Use your calculator to simplify the left side
    
    
• Base 10 (aka Common) logarithms.
  • Find the base 10 logarithm of each side:
    
  • Use the property to move the exponent out in front:
    
  • Divide both sides by :
    
  • Substitute the approximation 2.7182818284590451 for e. (Feel free to round this off some.):
    
  
  
  • Use your calculator to simplify the left side
    
    
    

So it will be about 200 years before the 700 pounds will be reduced to 150 pounds. (Note that we get the same answer using either natural or base 10 logarithms. (The very slight difference is due to using an approximation for e.))

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ln(x) = 11

you want to find x?

In general,

ln(x) = y if and only if e^y = x

if y = 11, this means that:

e^11 = x

e^11 = 59874.14171 which means that:

x = 59874.14171

This means that:

ln(x) = 11 becomes:

ln(59874.14171) = 11 which becomes:

11 = 11 confirming the value of x is good.

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Formula for continuous compounding is:

FV = PA * e^(rt)

FV = future value
PA = present amount
e = scientific constant of 2.718281828.....
r = annual interest rate
t = time in years

In your equation,

FV = 2
PA = 1
e = 2.71828182818.....
r = r
t = 12

Equation becomes:

2 = 1 * e^(12r)

This becomes:

2 = e^(12r)

Take natural log of both sides (natural log is log to the base of e)

ln(2) = ln(e^(12r)

By laws of logarithms, this becomes:

ln(2) = 12r*ln(e)

ln(e) = 1

Equation becomes:

ln(2) = 12r

Divide both sides by 12 to get:

ln(2)/12 = r

Solve for r to get:

r = .057762265 * 100% = 5.7762265%

Using continuous compounding, an annual interest rate of 5.7762265% will double your money in 12 years.

Confirm by plugging into the original equation.

2 = 1 * e^(.057762265*12)

Simplify to get:

2 = 2 confirming the value of r is good.

You might try some of the LESSONS FROM ALGEBRA.COM if you have the time. There's a few in the financial area that might help you.

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Strictly speaking, from the way you worded your question, there are no real values of for which has purely imaginary roots. That is because the coefficient on the 1st degree term is non-zero and rational, hence any roots of the given equation must have a rational (and therefore real) component. There is a range of values for for which will have a pair of complex roots of the form where and .

Remember the discriminant:

If then the roots are a conjugate pair of complex numbers of the form . Only in the case where is the real part of the complex number equal to zero and the complex number purely imaginary.

So:



And for , , hence or .

Actually it is possible to have purely imaginary roots if you allow , but somehow I don't think that is what you were talking about when you asked the question.

John

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log base 5 x=3


x = 125

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Some keys to understanding how to do this problem:

• Using percents in formulas or functions is not very practical. So, since 75% is 3/4, we will use 3/4 instead of 75%. (We could also use 0.75 instead of 75%.)
• When problems say "75% of" some number (or "3/4 of") some number, that "of" translates into a multiplication. The beer-lambert-bougar law says that the intensity of light is some percent of the intensity of the light 1 meter above. So to find the intensity of light in your lake, we will multiply the intensity of light above by 75% (or 3/4).

So to figure out the answers to your problem, let's make up a number for the intensity of light just as it hits the water (i.e. the depth is zero):

• Let's say the intensity at depth 0 is 100.
• The intensity at a depth of 1 meter would be 3/4 of intensity of the light at 0 meters: .
• At 2 meters depth the intensity would be 3/4 of the intensity at 1 meters depth:
• At 3 meters depth the intensity would be 3/4 of the intensity at 2 meters depth:
• etc.

a) Explain why intensity I is an exponential function of depth d in meters
We can see that the intensity is calculated, in part, by the repeated multiplication of 3/4. Our formula is an exponential function because we will use an exponent to represent the repeated multiplication of 3/4.

b) Use a formula to express intensity I as an exponential function of d. (use I0 to denote the initial intensity.)
From the example above we can see where the initial intensity (intensity at depth 0) fits in the formula:

where I is the intensity at d meters, is the initial intensity and d is the depth in meters.

c) Explain in practical terms the meaning of .
It is the intensity at depth 0 meters

d) At what depth will the intensity of light be one tenth of the intensity of light striking the surface?
One tenth of the intensity of light striking the surface is (Remember that "a fraction of a number" means that fraction times that number.) So the equation we need to solve is:

First we'll divide by

On both sides the 's cancel leaving:

To solve for d, when it is the exponent like this, we will use logarithms:

The left side is -1 because 1/10 is . (If this is not obvious, use your calculator. So now we have:

On the right side we can use the property of logarithms, , to move the exponent from the argument out in front of the logarithm giving us:

And now we can divide both sides by :

On the right side the 's cancel leaving:

Using our calculator on the remaining log:

And using the calculator to divide we get:

So at very close to 8 meters the intensity of light will be 1/10 of what it was at the surface.

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For problems like yours, where the variable is the base or in the argument of the logarithm, the key is to rewrite the logarithmic equation in exponential form. To do this we need to know that is equivalent to .


Rewriting this in exponential form we get:

The equation is now solvable. We just square both sides:

which simplifies to



Rewriting this in exponential form we get:

If you are a computer geek you may recognize that . Otherwise we find the 10th root of each side:

The left side is the same as which we can find using a calculator. The right side simplifies to . So now we have:

Solving this we get
or
And, since x is the base of a logarithm, we must reject x = -2. So the only solution is x = 2.

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Your question is not cleared as typed.
Do you mean 42x^2 - 1 = 8^(x)?
In other words, is 8 raised to a variable?
Get back to me before we can go on with this question.

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Factoring out the minus -1 was a good idea. But, as you found, separating log(-1) was not. With , the transformation from is:

• A translation to the right of 8
• A reflection in the y-axis

Another way could be to factor out -1:

Factor the -1 into 10*(-1/10):

Separate out the factor of 10:

Since :

The transformations from would be:

• A translation to the right of 8
• A translation up of 1
• A reflection in the y-axis (because of the "-")
• A horizontal spreading or stretching by a factor of 10 (Stretching, not compression, because of the fraction)

I don't know if this second one is "very different" enough.

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.04 / 7 * 100% = .571428571%

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P=C(1+r/n)^nt for compounding interest what does the ^ stand for and how would you calculate it if:
C= 3150
r=2
n=365
t=1
.

Plugging in our given values:





.
That's $3213.63

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Replacing x by (x-2) in the right side causes a shift of 2 units RIGHT.

Multiplying the right side by 3 causes a vertical STRETCH of 3.

Answer:

shift 2 units right vertical stretch of 3

Edwin