# small question

• December 3rd 2008, 10:13 AM
larasanut
small question
what do i use to get the bases the same.. 8^x = 1/32^x-2
• December 3rd 2008, 10:33 AM
Soroban
Hello, larasanut!

Quote:

What do i use to get the bases the same? . . . $8^x \:=\: \frac{1}{32^{x-2}}$

Note that: . $8 \,=\,2^3\:\text{ and }\:32 \,=\,2^5$

So we have: . $\left(2^3\right)^x \;=\;\frac{1}{(2^5)^{x-2}} \;=\;\frac{1}{2^{5x-10}} \;=\;2^{-(5x-10)}$

Hence: . $2^{3x} \;=\;2^{-5x+10}$ . . . Got it?

• December 3rd 2008, 10:43 AM
larasanut
YUP THANKS