# Solving $\;5^{2x}-4\cdot 5^x=12$

I need to solve $\quad\displaystyle 5^{2x}-4\cdot 5^x=12$.

I've only gotten this far: $\quad \displaystyle 5^{2x}-20^x=12.$

I don't know what to do next.

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Let $5^x=y$ in the initial equation (your second line is false). – L. F. Jan 28 '13 at 15:29
$4\times 5^x\neq 20^x$ – uforoboa Jan 28 '13 at 15:29

HINT Setting $z=5^x$ gives the equation $z^2-4z=12$. Do you see how to proceed?

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Please note that $4\cdot5^x = 4(5^x)\neq 20^x$

If we let $y=5^x$, then $$(5^x)^2-4(5^x) =12 \;\implies \;y^2 - 4y = 12 \;\implies y^2 - 4y - 12 = (y-6)(y+2) = 0$$

$\implies \; y = 6\;$ or $\;y = -2$

So $y = 5^x = 6\;$ or $\; y = 5^x = -2$

Can you take it from here?

You can omit $\;y= 5^x = -2\;$ as a solution if we are constraining ourselves to real solutions...

So it suffices to solve for $x$ given $\;5^x = 6$.

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I dont feel like these kind of equations. + – Babak S. Jan 28 '13 at 19:40

just let 5^x=y and solve the quadratic equation.

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First of all, $4\cdot5^x \neq 20^x$. Take for example $x=2$, this would imply $100 = 400$.

To solve this you should notice that $5^{2x}$ = $(5^x)^2$. Then if you substitute some "dummy variable" $u$ for $5^x$ you get a quadratic $$u^2 - 4u -12 = 0$$

You know how to solve this quadratic, by factoring to get $(u-6)(u+2)=0$ and having two solutions $u=6,-2$.

You take these two solutions and substitute back in the $5^x$, because this $u$ doesn't mean anything other than a way to solve the problem. And you get two equations: $$5^x = 6\\5^x = -2$$ With two solutons, one for each problem: $$x = \log_5 6\\x=\log_5(-2)$$

But the logarithm of a negative number doesn't make sense without imaginary numbers, so your only real solution is $$x=\log_5 6$$

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