Question

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

Answer 1

`X = 8^(1/8)=1.297(cos( (2kpi)/8)+isin((2kpi)/8)), k=0, 1, 2,.,7`
The two real solutions are `+-1.297`, near.y, for k = 0 and k = 4..

Explanation:

Use `m log a = log a^m and log( (a^m)^n)=log a^(mn)`.

`log_3 8=log_3((X^2)^4)=log_3X^8`
`X^8=8`
`X=8^(1/8)=1.297(cos (2kpi)+isin(2kpi))^(1/8`, k=intege (including 0),
Now, `X=1.297(cos( (2kpi)/8)+isin((2kpi)/8))`, k=0, 1, 2, ...7, for the eight values repeated in a cycle.

The two real solutions are `+-1.297`, near.y, for k = 0 and k = 4.#