# Solving a logarithmic equation

• xxwinexx
Remember your exponent and logarithm rules (they're pretty similar): If a, b, and c are real numbers, then:(ab)c = abcandln(a) + ln(b) = ln(ab).The latter is derived from the fact that eaeb = ea+b.Indeed. You can check it by replacing x with ln(5) in the original equation. Since both sides are positive (the natural logarithm has only a positive domain), it is also the only real solution.

(e^x)^2-5(e^x)=0

## Homework Equations

I've reviewed my logarithmic rules, but I cannot get to the right solution. I should be trying to get one e^x on each side so that I can take the ln of both sides and end up with a simple algebraic process to solve.

## The Attempt at a Solution

This is as far as I get without running into my problem:
(e^x)^2=5(e^x)

I guess here I could take the ln of both sides, but I guess I'm confused about what happens to the x^2, and to the 5.

Edit:

Ahh, Thanks for the reminder. Don't want to bump this any more.

Took ln of both sides, ended with:

2x=ln5+x

x=ln5

Is that the correct way to that solution?

Last edited:
Remember your exponent and logarithm rules (they're pretty similar): If a, b, and c are real numbers, then:
(ab)c = abc
and
ln(a) + ln(b) = ln(ab).
The latter is derived from the fact that eaeb = ea+b.

Indeed. You can check it by replacing x with ln(5) in the original equation. Since both sides are positive (the natural logarithm has only a positive domain), it is also the only real solution.

xxwinexx said:

(e^x)^2-5(e^x)=0

## Homework Equations

I've reviewed my logarithmic rules, but I cannot get to the right solution. I should be trying to get one e^x on each side so that I can take the ln of both sides and end up with a simple algebraic process to solve.
If you let y= ex, then your equation is y2- 5y= y(y- 5)= 0. Can you solve that?

## The Attempt at a Solution

This is as far as I get without running into my problem:
(e^x)^2=5(e^x)

I guess here I could take the ln of both sides, but I guess I'm confused about what happens to the x^2, and to the 5.
?? There is NO x^2. There is (e^x)^2.

Edit:

Ahh, Thanks for the reminder. Don't want to bump this any more.

Took ln of both sides, ended with:

2x=ln5+x

x=ln5

Is that the correct way to that solution?