Question 404283
{{{2*log(8,4) - (1/3)*log(8,8)}}}
Use the general rule:
{{{a*log(b) = log(b^a)}}}
{{{log(8,4^2) - log(8,8^(1/2))}}}
Now use the rule:
{{{log(a) - log(b) = log(a/b)}}} (logs with the same base)
{{{log(8, (16/sqrt(8)))}}}
{{{log(8, (8/sqrt(2)))}}}
Now you can use the reverse of the above rule:
{{{log(a/b) = log(a) - log(b)}}}
{{{log(8,8) - log(8,sqrt(2))}}}
Rule:
{{{log(a,a) = 1}}}
{{{1 - log(8,sqrt(2))}}}
Now, can I get {{{sqrt(2)}}} in terms of {{{8}}}?
I know {{{8^(1/3) = 2}}}
so, {{{sqrt(8^(1/3)) = sqrt(2)}}}
This is the same as:
{{{8^(1/6) = sqrt(2)}}}
therefore
{{{1 - log(8,sqrt(2)) = 1 - log(8,8^(1/6))}}}
rule:
{{{log(a,a^b) = b}}}
{{{1 - 1/6 = 5/6}}} answer
Unless I goofed