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!

Re: Solving Logarithmic/Exponential Equations

Quote:

Originally Posted by

**radh** 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?

Re: Solving Logarithmic/Exponential Equations

It's 25^(x-1) and 36^(x-9). I apologize for the confusion.

Re: Solving Logarithmic/Exponential Equations

You are correct up to $\displaystyle \displaystyle \begin{align*} 2x-2 = 12x \end{align*}$, so

$\displaystyle \displaystyle \begin{align*} 2x - 2 &= 12x \\ -2 &= 10x \\ -\frac{2}{10} &= x \\ x &= -\frac{1}{5} \end{align*}$