what do i use to get the bases the same.. 8^x = 1/32^x-2
Hello, larasanut!
What do i use to get the bases the same? . . . $\displaystyle 8^x \:=\: \frac{1}{32^{x-2}}$
Note that: .$\displaystyle 8 \,=\,2^3\:\text{ and }\:32 \,=\,2^5$
So we have: .$\displaystyle \left(2^3\right)^x \;=\;\frac{1}{(2^5)^{x-2}} \;=\;\frac{1}{2^{5x-10}} \;=\;2^{-(5x-10)}$
Hence: .$\displaystyle 2^{3x} \;=\;2^{-5x+10}$ . . . Got it?