# Difficulty solving exponential equation

the question is little silly, but I have difficulty in math so, how to solve this exponential equation ?

$$\ 4^x - 4^1 + 2^x - 2^1 = \frac{5}{16}.$$

How do I manage to solve it ?

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What if you substitute $x=\log_2 t$? – Antonio Vargas Mar 19 '12 at 2:03

First, $4^1=4$ and $2^1=2$, so you can write it as $4^x+2^x-6=\frac5{16}$. Next, $4=2^2$, so $4^x=(2^2)^x=2^{2x}=(2^x)^2$. For convenience let $y=2^x$; then you can rewrite the equation yet again as $y^2+y-6=\frac5{16}$, or $y^2+y-\frac{101}{16}=0$. This is a straightforward quadratic equation in $y$; solve it to find $y$, then take logarithms to get $\log y=x\log 2$, so that $x=\frac{\log y}{\log 2}$.