# How do you solve 2^x=64?

Oct 16, 2016

I got $x = 6$

#### Explanation:

We can write it as:
${2}^{x} = {2}^{6}$
so that $x = 6$

Oct 16, 2016

$x = 6$

#### Explanation:

${2}^{x} = 64$

Just in case your teacher wants you to solve this the fancy way...

$\log {2}^{x} = \log 64 \textcolor{w h i t e}{a a a}$Take the log of both sides

$x \log 2 = \log 64 \textcolor{w h i t e}{a a a}$Use the log rule $\log {x}^{a} = a \log x$

$\frac{x \log 2}{\log} 2 = \log \frac{64}{\log} 2 \textcolor{w h i t e}{a a a}$Divide both sides by log2

$x = 6$

Or, if you're not allowed to use a calculator...

${\log}_{2} {2}^{x} = {\log}_{2} 64$

$x {\log}_{2} 2 = {\log}_{2} 64$

$\frac{x {\log}_{2} 2}{\log} _ 2 2 = {\log}_{2} \frac{64}{\log} _ 2 2$

${\log}_{\textcolor{red}{2}} \textcolor{v i o \le t}{64} = \textcolor{b l u e}{6}$ because the answer to a log problem is the exponent $\textcolor{b l u e}{6}$ that will make the base $\textcolor{red}{2}$ equal the argument $\textcolor{v i o \le t}{64}$. In other words, ${\textcolor{red}{2}}^{\textcolor{b l u e}{6}} = \textcolor{v i o \le t}{64}$. Similarly, ${\log}_{\textcolor{red}{2}} \textcolor{v i o \le t}{2} = \textcolor{b l u e}{1}$ because ${\textcolor{red}{2}}^{\textcolor{b l u e}{1}} = \textcolor{v i o \le t}{2}$. But if you knew that, you could solve this problem without all the fancy math!

$x = \frac{6}{1}$

$x = 6$