Exponential using two different bases.

8^{x-1 }= 9^{x+3 }Solve for x.

My attempt:

I'm guessing since they're both equal, if I multiply them it will just be the same as squaring one side.

We have (8^{x-1})(9^{x+3})=72^{x^2+2x-3 }

ANNNNND I don't know what to do next. I tried equating x^{2}+2x-3 = 2 since 72^{x^2+2x-3 }= a^{2} where a is the base. Solving for x^{2}+2x-5=0, I got:

$\displaystyle x = \ -1\pm\sqrt{6}$

How do I solve this? I need to find the base.

Re: Exponential using two different bases.

Hey RPSGCC733.

Recall that you can calculate the log in any base by log_b(x) = log_c(x)/log_c(b) where log_c(x) is the logarithm in base c.

This means that you can calculate log_9(x) in terms of log_8(x) and vice versa.