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#1

$Old$ December 3rd, 2008, 10:13 AM

larasanut $larasanut is offline$

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$Default$ small question

what do i use to get the bases the same.. 8^x = 1/32^x-2

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#2

$Old$ December 3rd, 2008, 10:33 AM

Soroban $Soroban is offline$

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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?

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#3

$Old$ December 3rd, 2008, 10:43 AM

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