# How do you solve log _ 6 ( log _ 2 (5.5x)) = 1?

Mar 4, 2018

$x = \frac{128}{11} = 11. \overline{63}$

#### Explanation:

We begin by raising both sides as a power of $6$:

${\cancel{6}}^{\cancel{{\log}_{6}} \left({\log}_{2} \left(5.5 x\right)\right)} = {6}^{1}$

${\log}_{2} \left(5.5 x\right) = 6$

Then we raise both sides as powers of $2$:

${\cancel{2}}^{\cancel{{\log}_{2}} \left(5.5 x\right)} = {2}^{6}$

$5.5 x = 64$

$\frac{\cancel{5.5} x}{\cancel{5.5}} = \frac{64}{5.5}$

$x = \frac{128}{11} = 11. \overline{63}$

Mar 4, 2018

$x = \frac{128}{11} \approx 11.64$

#### Explanation:

Recall that ${\log}_{b} a = m \iff {b}^{m} = a \ldots \ldots \ldots . \left(\lambda\right)$.

Let, ${\log}_{2} \left(5.5 x\right) = t$.

Then, ${\log}_{6} \left({\log}_{2} \left(5.5 x\right)\right) = 1 \Rightarrow {\log}_{6} \left(t\right) = 1$.

$\Rightarrow {6}^{1} = t \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left[\because , \left(\lambda\right)\right]$.

$\Rightarrow t = {\log}_{2} \left(5.5 x\right) = 6$.

$\therefore \text{By } \left(\lambda\right) , {2}^{6} = 5.5 x$.

$\therefore 5.5 x = 64$.

$\Rightarrow x = \frac{64}{5.5} = \frac{128}{11} \approx 11.64$