# How do you condense 6log_2 (2/3)+2log_2(1/6)-4log_2(2/9)?

Mar 23, 2016

The expression can be condensed to $0$.

#### Explanation:

We can use the following logarithmic laws:

$\left[1\right] \text{ } {\log}_{a} x + {\log}_{a} y = {\log}_{a} \left(x \cdot y\right)$

$\left[2\right] \text{ } {\log}_{a} x - {\log}_{a} y = {\log}_{a} \left(\frac{x}{y}\right)$

$\left[3\right] \text{ } r \cdot {\log}_{a} x = {\log}_{a} \left({x}^{r}\right)$

As all logarithmic expressions have the same basis, all those laws can be applied without any further preparation.

Thus, we have:

$6 {\log}_{2} \left(\frac{2}{3}\right) + 2 {\log}_{2} \left(\frac{1}{6}\right) - 4 {\log}_{2} \left(\frac{2}{9}\right)$

$\stackrel{\text{[3] }}{=} {\log}_{2} \left({\left(\frac{2}{3}\right)}^{6}\right) + {\log}_{2} \left({\left(\frac{1}{6}\right)}^{2}\right) - {\log}_{2} \left({\left(\frac{2}{9}\right)}^{4}\right)$

$\stackrel{\text{[1] }}{=} {\log}_{2} \left({\left(\frac{2}{3}\right)}^{6} \cdot {\left(\frac{1}{6}\right)}^{2}\right) - {\log}_{2} \left({\left(\frac{2}{9}\right)}^{4}\right)$

$\stackrel{\text{[2] }}{=} {\log}_{2} \left(\frac{{\left(\frac{2}{3}\right)}^{6} \cdot {\left(\frac{1}{6}\right)}^{2}}{\frac{2}{9}} ^ 4\right)$

Now, the expression is condensed. However, it can still be simplified:

$\frac{{\left(\frac{2}{3}\right)}^{6} \cdot {\left(\frac{1}{6}\right)}^{2}}{\frac{2}{9}} ^ 4 = \frac{{2}^{6} / {3}^{6} \cdot \frac{1}{6} ^ 2}{{2}^{4} / {9}^{4}}$

... dividing by a fraction is the same thing as multiplying by the reciprocal...

$\text{ } = \frac{{2}^{6} \cdot {9}^{4}}{{3}^{6} \cdot {6}^{2} \cdot {2}^{4}}$

$\text{ } = \frac{{2}^{6} \cdot {\left({3}^{2}\right)}^{4}}{{3}^{6} \cdot {\left(2 \cdot 3\right)}^{2} \cdot {2}^{4}}$

... use the power rule ${\left({a}^{m}\right)}^{n} = {a}^{m \cdot n}$:

$\text{ } = \frac{{2}^{6} \cdot {3}^{8}}{{3}^{6} \cdot {2}^{2} \cdot {3}^{2} \cdot {2}^{4}}$

... use the power rule ${a}^{m} \cdot {a}^{n} = {a}^{m + n}$:

$\text{ } = \frac{{2}^{6} \cdot {3}^{8}}{\textcolor{b l u e}{{3}^{6}} \cdot \textcolor{\mathmr{and} a n \ge}{{2}^{2}} \cdot \textcolor{b l u e}{{3}^{2}} \cdot \textcolor{\mathmr{and} a n \ge}{{2}^{4}}}$

$\text{ } = \frac{{2}^{6} \cdot {3}^{8}}{\textcolor{b l u e}{{3}^{8}} \cdot \textcolor{\mathmr{and} a n \ge}{{2}^{6}}}$

... the numerator and the denominator can be cancelled completely...

$\text{ } = 1$

Thus, in total, we have:

${\log}_{2} \left(\frac{{\left(\frac{2}{3}\right)}^{6} \cdot {\left(\frac{1}{6}\right)}^{2}}{\frac{2}{9}} ^ 4\right) = {\log}_{2} \left(1\right) = 0$,

as for any basis $a$, it holds ${\log}_{a} \left(1\right) = 0$ .