# How do you solve log_x 2 + log_2 x = 3 ?

Dec 3, 2017

$x = {2}^{\frac{3 \pm \sqrt{5}}{2}}$

#### Explanation:

First we will convert ${\log}_{x} \left(2\right)$ to ${\log}_{2}$. We will use the following log property:
${\log}_{b} \left(a\right) = {\log}_{x} \frac{a}{\log} _ x \left(b\right)$

In our equation, this becomes:
${\log}_{x} \left(2\right) + {\log}_{2} \left(x\right) = {\log}_{2} \frac{2}{\log} _ 2 \left(x\right) + {\log}_{2} \left(x\right) = \frac{1}{\log} _ 2 \left(x\right) + {\log}_{2} \left(x\right)$

This is a quadratic, but to make it easier to understand I will introduce a substitution so $u = {\log}_{2} \left(x\right)$:
$\frac{1}{\log} _ 2 \left(x\right) + {\log}_{2} \left(x\right) = 3$

$\frac{1}{u} + u = 3$

Multiply by $u$ on both sides:
$1 + {u}^{2} = 3 u$

${u}^{2} - 3 u + 1 = 0$

$u = \frac{3 \pm \sqrt{5}}{2}$

Now we resubstitute, so we have
${\log}_{2} \left(x\right) = \frac{3 \pm \sqrt{5}}{2}$

We put both sides as powers of $2$:
${\cancel{2}}^{\cancel{{\log}_{2}} \left(x\right)} = {2}^{\frac{3 \pm \sqrt{5}}{2}}$

$x = {2}^{\frac{3 \pm \sqrt{5}}{2}}$

Dec 3, 2017

x = 2^(1/2(3+-sqrt(5))

#### Explanation:

Given:

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

By the change of base formula, we have:

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

Letting $t = \log \frac{x}{\log} 2 = {\log}_{2} x$, that becomes:

$\frac{1}{t} + t = 3$

Mutliplying through by $t$ and rearranging a bit:

$0 = 4 \left({t}^{2} - 3 t + 1\right)$

$\textcolor{w h i t e}{0} = 4 {t}^{2} - 12 t + 4$

$\textcolor{w h i t e}{0} = {\left(2 t\right)}^{2} - 2 \left(2 t\right) \left(3\right) + {3}^{2} - 5$

$\textcolor{w h i t e}{0} = {\left(2 t - 3\right)}^{2} - {\left(\sqrt{5}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 t - 3\right) - \sqrt{5}\right) \left(\left(2 t - 3\right) + \sqrt{5}\right)$

$\textcolor{w h i t e}{0} = \left(2 t - 3 - \sqrt{5}\right) \left(2 t - 3 + \sqrt{5}\right)$

So:

$2 t = 3 \pm \sqrt{5}$

That is:

$2 {\log}_{2} x = 3 \pm \sqrt{5}$

So:

x = 2^(1/2(3+-sqrt(5))