How do you solve 2log_6 4-1/4log_6 16=log_6 x?

Mar 15, 2018

$x = 8$

Explanation:

We have: $2 {\log}_{6} \left(4\right) - \frac{1}{4} {\log}_{6} \left(16\right) = {\log}_{6} \left(x\right)$

Let's begin this by expressing $- \frac{1}{4} {\log}_{6} \left(16\right)$ in terms of an argument of $4$:

$R i g h t a r r o w 2 {\log}_{6} \left(4\right) - \frac{1}{4} {\log}_{6} \left({4}^{2}\right) = {\log}_{6} \left(x\right)$

$R i g h t a r r o w 2 {\log}_{6} \left(4\right) - \frac{2}{4} {\log}_{6} \left(4\right) = {\log}_{6} \left(x\right)$

$R i g h t a r r o w 2 {\log}_{6} \left(4\right) - \frac{1}{2} {\log}_{6} \left(4\right) = {\log}_{6} \left(x\right)$

Then, we can subtract the like terms on the left-hand side of the equation:

$R i g h t a r r o w \left(2 - \frac{1}{2}\right) {\log}_{6} \left(4\right) = {\log}_{6} \left(x\right)$

$R i g h t a r r o w \frac{3}{2} {\log}_{6} \left(4\right) = {\log}_{6} \left(x\right)$

$R i g h t a r r o w {\log}_{6} \left({4}^{\frac{3}{2}}\right) = {\log}_{6} \left(x\right)$

Now, both sides of the equation are in terms of the logarithm of base $6$.

We can cancel these logarithms out by exponentiating both sides by $6$:

$R i g h t a r r o w {6}^{{\log}_{6} \left({4}^{\frac{3}{2}}\right)} = {6}^{{\log}_{6} \left(x\right)}$

$R i g h t a r r o w {4}^{\frac{3}{2}} = x$

$\therefore x = 8$