# How do you solve 4^(2x-5)=32?

Mar 9, 2018

Real solution $x = \frac{15}{4}$

Complex solutions $x = \frac{15}{4} + \frac{k \pi}{2 \ln 2} i$ for any integer $k$

#### Explanation:

Note that $4 = {2}^{2}$ and $32 = {2}^{5}$.

So we have:

${2}^{4 x - 10} = {\left({2}^{2}\right)}^{2 x - 5} = {4}^{2 x - 5} = 32 = {2}^{5}$

The function ${2}^{x}$ is one to one as a real valued function from $\mathbb{R}$ to $\left(0 , \infty\right)$.

Hence the only real solution is given by:

$4 x - 10 = 5$

Hence:

$x = \frac{15}{4}$

If we are interested in other complex solutions, note that:

${2}^{\frac{2 k \pi i}{\ln} 2} = {e}^{2 k \pi i} = {\left({e}^{2 \pi i}\right)}^{k} = {1}^{k} = 1 \text{ }$ for any integer $k$

Hence the given equation has solutions given by:

$4 x - 10 = 5 + \frac{2 k \pi i}{\ln} 2$

Hence:

$x = \frac{15}{4} + \frac{k \pi}{2 \ln 2} i$

Mar 9, 2018

$\frac{15}{4}$

#### Explanation:

In order to solve, you can manipulate the bases of the exponents to find a value for x.

In this question, you can manipulate the base of 4 to be ${2}^{2}$ and 32 to be ${2}^{5}$. It should look like this:
${2}^{2 \left(2 x - 5\right)} = {2}^{5}$

Now that the bases equivalent, you can take a logarithm with a base of 2 to both sides of the equation to cancel out the exponents.
${\log}_{2} \left({2}^{2 \left(2 x - 5\right)}\right) = {\log}_{2} \left({2}^{5}\right)$

${\log}_{2} \left(2\right) = 1$ so you can cancel out the log functions on both sides. This gives you:
$2 \left(2 x - 5\right) = 5$

Now, just solve like a normal algebraic equation.
$4 x - 10 = 5$
$4 x = 15$
$x = \frac{15}{4}$