# How can I apply log laws here?

Solve for $x$: $$2^{2x+1} - (17)2^x + 8 = 0$$

I have the answers: -1, 3

I tried a few different transformations, but couldn't get a clear answer. I suspect that I am overlooking a property of log that would be useful.

Edit: using a u substitution for $2^x$, I was able to factor the quadratic into:

$(2u-1)(u-8)$

$2^x = 1/2$

$2^x = 8$

$x = -1, 3$

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Don't bother with log at first. The real question is, what is $2^{2x+1}$ in terms of $2^x$? I.e. if you have the latter, what do you do to it to get the former? That way you can say $u=2^x$ is its own variable, rewrite the equation entirely with $u$, solve for $u$, and then take the base-2 logarithm to find $x$.

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Note that $2^{2x+1} = 2^{2x}\times 2 = 2\times(2^x)^2$.

So if we let $y=2^x$, then you can rewrite the equation as $$2y^2 - 17y + 8 = 0.$$

This can be solved using the quadratic formula. Once you know the value of $y=2^x$, then you can use logarithms to get the value of $x$. However, applying the logarithm to the whole expression on the left of the equal sign of your original equation will not simplify things.

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This problem should be closed because $(2y-1)(y-8)=0 \Rightarrow x=-1, 3$ are the solutions. –  Kirthi Raman Mar 11 '12 at 22:59
@KirthiRaman: I'm sorry... why does that imply that "the problem should be closed"? –  Arturo Magidin Mar 11 '12 at 23:06