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

1 Answer
Apr 18, 2016

Answer:

# 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.#