Alex P.

asked • 11/11/14

How to solve 8(1-x)=1/4^(x+2)

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Mark W. answered • 11/11/14

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Alex, is that 8^(1-x)?  If so, it is much easier.  What math class is this for?
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Mark W.

Assuming it is 8^(1-x)=(1/4)^(x+2), then we can substitute:
8=2^3, and (1/4)=2^(-2)
So, rewriting the equation, we have
(2^3)^(1-x)=(2^(-2))^(x+2)
This then means:
2^(3*(1-x))=2^(-2*(x+2))
This is then:
2^(3-3x)=2^(-2x-4)

We then take the logarithm to the base 2 of both sides, and just set the exponents equal to each other:
3-3x=-2x-4
We add 3x and 4 to both sides.  On the left, the x's cancel, on the right, the 4's cancel:
3+4=3x-2x
Simplifying, we have 7=x.
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11/11/14

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