# How do you solve 5 (2^(3x)) = 8?

May 7, 2018

$x = \setminus \frac{{\log}_{2} \left(\setminus \frac{8}{5}\right)}{3}$

#### Explanation:

We must isolate the $x$: first of all, get rid of the $5$, dividing both sides by $5$:

${2}^{3 x} = \setminus \frac{8}{5}$

Now, take logarithm (base $2$) to both sides:

$3 x = {\log}_{2} \left(\setminus \frac{8}{5}\right)$

Finally, divide both sides by $3$:

$x = \setminus \frac{{\log}_{2} \left(\setminus \frac{8}{5}\right)}{3}$

If you prefer, you can use the rule $\log \left(\frac{a}{b}\right) = \log \left(a\right) - \log \left(b\right)$ to write

${\log}_{2} \left(\setminus \frac{8}{5}\right) = {\log}_{2} \left(8\right) - {\log}_{2} \left(5\right) = 3 - {\log}_{2} \left(5\right)$

And the answer becomes

$x = 1 - \setminus \frac{{\log}_{2} \left(5\right)}{3}$

May 7, 2018

$x = \frac{\log 1.6}{3 \log \left(2\right)} = \frac{\ln 1.6}{3 \ln \left(2\right)} \approx .226$

#### Explanation:

Given: $5 \left({2}^{3 x}\right) = 8$

First divide by $5 : \text{ } {2}^{3 x} = \frac{8}{5} = 1.6$

Log base 2 both sides: $\text{ } {\log}_{2} {2}^{3 x} = {\log}_{2} 1.6$

Use the logarithmic property: ${\log}_{b} {b}^{x} = x$

$3 x = {\log}_{2} 1.6$

$x = \frac{{\log}_{2} 1.6}{3} = \frac{1}{3} \left({\log}_{2} 1.6\right)$

Use the change of base formula to convert to either log base 10 or the natural log: ${\log}_{b} x = \frac{\log x}{\log 2} = \frac{\ln x}{\ln 2}$

$x = \frac{1}{3} \frac{\log 1.6}{\log \left(2\right)} = \frac{\log 1.6}{3 \log \left(2\right)} \approx .226$

May 7, 2018

$x = 1 - \frac{1}{3} {\log}_{2} \left(5\right)$

#### Explanation:

You need to work with exponents here.

$5 \cdot {2}^{3 x} = 8 = {2}^{3}$
That means ${2}^{3 x} = {2}^{3} / 5$
Divide with ${2}^{3}$ on both sides: ${2}^{3 \left(x - 1\right)} = \frac{1}{5}$

Take ${\log}_{2}$ on both sides:

$3 \left(x - 1\right) = - {\log}_{2} \left(5\right)$

or $x = 1 - \frac{1}{3} {\log}_{2} \left(5\right)$

Check:
5*2^(3(1-1/3log_2(5)) =5*2^(3-log_2(5)=$5 \cdot \frac{8}{5} = 8$