Using logarithms to solve for x

[xxMsJojoxx]

Why is the correct answer x=ln(120)?

I got:
=132
=132
x=ln(132)

Please explain?

 

[jonah2.0]

Beer soaked hint follows.

View attachment 22096
Why is the correct answer x=ln(120)?

I got:
View attachment 22097 =132
View attachment 22098=132
x=ln(132)

Please explain?

[MATH](e^x)^2-e^x-132=0[/MATH]Can you factor it out?

 

[Deleted member 4993]

View attachment 22096
Why is the correct answer x=ln(120)?

I got:
View attachment 22097 =132
View attachment 22098=132
x=ln(132)

Please explain?

It seems that you have assumed:

e^2x = e^x + e^x = 2 * e^x ................INCORRECT

The CORRECT expression would be:

e^2x = e^x * e^x = (e^x)² ............................ edited [LaTex]

Follow the method shown in response #2 above and use solution for quadratic equation.

 

[pka]

View attachment 22096
Why is the correct answer x=ln(120)?

\(e^{2x}-e^x-132=(e^x-12)(e^x+11)\) SEE HERE
Also HERE

 

[JeffM]

[MATH]e^{2x} \div e^x = e^{2x} * e^{-x} = e^{(2x - x)} = e^x.[/MATH]
You are mixing up the rule for when you subtract exponents.

Forget e for a moment.

[MATH]2^6 - 2^3 = 64 - 8 = 56 \ne 8 = 2^3.[/MATH]

 

[Steven G]

2e^x - e^x= e^x, not e^2x - e^x = e^x