# How do you solve 4^x-2^(x+1)=3?

Apr 27, 2016

$x = \log \frac{3}{\log} 2$..

#### Explanation:

Use ${a}^{m n} = {\left({a}^{m}\right)}^{n} = {\left({a}^{n}\right)}^{m} \mathmr{and} {a}^{m + n} = {a}^{m} {a}^{n}$

${4}^{x} = {\left({2}^{2}\right)}^{x} = {2}^{2 x} = {\left({2}^{x}\right)}^{2}$

The given equation is ${u}^{2} - 2 u - 3 = 0$, where $u = {2}^{x}$
The roots are u = ${2}^{x} = 3 \mathmr{and} - 1$. As ${2}^{x} > 0$, for all $x , - 1$ is inadmissible..

Now solve ${2}^{x} = 3$, by equating the logarithms.

$x = \log \frac{3}{\log} 2$ .