# How do you condense log_3 8 - 4log_3(X^2)?

Apr 18, 2016

$X = {8}^{\frac{1}{8}} = 1.297 \left(\cos \left(\frac{2 k \pi}{8}\right) + i \sin \left(\frac{2 k \pi}{8}\right)\right) , k = 0 , 1 , 2 , . , 7$
The two real solutions are $\pm 1.297$, near.y, for k = 0 and k = 4..

#### Explanation:

Use $m \log a = \log {a}^{m} \mathmr{and} \log \left({\left({a}^{m}\right)}^{n}\right) = \log {a}^{m n}$.

${\log}_{3} 8 = {\log}_{3} \left({\left({X}^{2}\right)}^{4}\right) = {\log}_{3} {X}^{8}$
${X}^{8} = 8$
X=8^(1/8)=1.297(cos (2kpi)+isin(2kpi))^(1/8, k=intege (including 0),
Now, $X = 1.297 \left(\cos \left(\frac{2 k \pi}{8}\right) + i \sin \left(\frac{2 k \pi}{8}\right)\right)$, k=0, 1, 2, ...7, for the eight values repeated in a cycle.

The two real solutions are $\pm 1.297$, near.y, for k = 0 and k = 4.#