Solving Logarithmic/Exponential Equations

• May 11th 2012, 11:22 PM
Solving Logarithmic/Exponential Equations
Hey everyone, I've tried these log problems, but to no avail.

25^(x-1) = 125^4x
I tried this:
(5^2)^x-1 = (5^3)^4x
5^(2x-2) = 5^12x
2x-2 = 12x
Here, I was confused. No substitution would make it work. (The instant math solver says -1/5)
Am I going at this the wrong way?

Thank you for the help!
• May 11th 2012, 11:28 PM
Prove It
Re: Solving Logarithmic/Exponential Equations
Quote:

Hey everyone, I've tried these log problems, but to no avail.

25^x-1 = 125^4x
I tried this:
(5^2)^x-1 = (5^3)^4x
5^2x-2 = 5^12x
2x-2 = 12x
Here, I was confused. No substitution would make it work. (The instant math solver says -1/5)
Am I going at this the wrong way?

Also, for clarification (in two examples), here's another problem:
36^x-9 = 6^2x
I tried this:
(6^2)^x-9 = (6)^2x
6^2x-18 = 6^2x
2x-18 = 2x
Which once again comes up with an impossible substitution.

Thank you for the help!

Is it \displaystyle \displaystyle \begin{align*} 25^x - 1 = 125^{4x} \end{align*} or \displaystyle \displaystyle \begin{align*} 25^{x - 1} = 125^{4x} \end{align*}, or something else entirely?
• May 11th 2012, 11:31 PM
You are correct up to \displaystyle \displaystyle \begin{align*} 2x-2 = 12x \end{align*}, so
\displaystyle \displaystyle \begin{align*} 2x - 2 &= 12x \\ -2 &= 10x \\ -\frac{2}{10} &= x \\ x &= -\frac{1}{5} \end{align*}