# How do you solve Log_2x + log_2(x+2) = log_2(6x+1)?

Apr 18, 2018

${x}_{1} = 2 + \sqrt{5} \mathmr{and} {x}_{2} = 2 - \sqrt{5}$

#### Explanation:

$L o {g}_{2} x + {\log}_{2} \left(x + 2\right) = {\log}_{2} \left(6 x + 1\right)$

Remember that $\log \left(a\right) + \log \left(b\right) = \log \left(a \cdot b\right)$

$L o {g}_{2} \left(x \cdot \left(x + 2\right)\right) = {\log}_{2} \left(6 x + 1\right)$
$x \cdot \left(x + 2\right) = 6 x + 1$
${x}^{2} + 2 x = 6 x + 1 | - 2 x$
${x}^{2} = 4 x + 1 | - 4 x - 1$
${x}^{2} - 4 x - 1 = 0$
${\left(x - 2\right)}^{2} - 1 - 4 = 0 | + 5$
${\left(x - 2\right)}^{2} = 5 | \sqrt{} | + 2$
$x = 2 \pm \sqrt{5}$
${x}_{1} = 2 + \sqrt{5} \mathmr{and} {x}_{2} = 2 - \sqrt{5}$
x_2<0 -> log(x_2)∈CC

Apr 18, 2018

$x = 2 + \sqrt{5}$

#### Explanation:

${\log}_{2} x + {\log}_{2} \left(x + 2\right) = {\log}_{2} \left(6 x + 1\right)$

color(green)(loga+logb=logab

${\log}_{2} x \left(x + 2\right) = {\log}_{2} \left(6 x + 1\right)$

By taking both sides for powers of 2 like this:

2^(log_2x(x+2))=2^(log_2(6x+1)

color(green)(a^(log_ax)=x

Thus

$x \left(x + 2\right) = 6 x + 1$

${x}^{2} + 2 x = 6 x + 1$

${x}^{2} - 4 x - 1 = 0$

$x = 2 + \sqrt{5}$

$\text{ OR }$

$x = 2 - \sqrt{5}$$\textcolor{red}{\text{ refused as it doesn't satisfy the equation}}$

Apr 18, 2018

$\textcolor{b l u e}{x = 2 + \sqrt{5}}$

#### Explanation:

By the laws of logarithms:

${\log}_{a} \left(b\right) + {\log}_{a} \left(c\right) = {\log}_{a} \left(b c\right)$

${\log}_{a} \left(b\right) = {\log}_{a} \left(c\right) \iff b = c$



${\log}_{2} \left(x\right) + {\log}_{2} \left(x + 2\right) = {\log}_{2} \left(6 x + 1\right)$

${\log}_{2} \left(x \left(x + 2\right)\right) = {\log}_{2} \left(6 x + 1\right)$

${\log}_{2} \left({x}^{2} + 2 x\right) = {\log}_{2} \left(6 x + 1\right)$

$\therefore$

${x}^{2} + 2 x = 6 x + 1$

${x}^{2} - 4 x - 1 = 0$

$x = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm 2 \sqrt{5}}{2} = 2 \pm \sqrt{5}$

We need to test this:

$x = 2 - \sqrt{5} < 0$

For:

${\log}_{2} \left(x\right)$ This is undefined for real mumbers.

So only the soloution:

$\textcolor{b l u e}{x = 2 + \sqrt{5}}$