Question 221046
<font size = 7 color = "red"><b>Edwin's full explanation:</b></font> 
<pre><font size = 4 color = "indigo"><b>

{{{4^(log(8,27))}}}

Since there is a {{{4}}} and an {{{8}}} involved, and
both are powers of {{{2}}}, let's try to get as much
as possible involving the number {{{2}}}:

Let's first work only on the exponent

{{{log(8,27)}}}

Use the change of base formula {{{log((OLD_BASE),(X))=log((NEW_BASE),(X))/log((NEW_BASE),(OLD_BASE))}}}

to change base {{{8}}} to base {{{2}}}

{{{log(2,27)/log(2,8)}}}

Write {{{27}}} as {{{3^3}}} and {{{8}}} as {{{2^3}}}

{{{log(2,3^3)/log(2,2^3)}}}

Use the rule {{{log((B),X^N)=N*log((B),X)}}} to rewrite
numerator and denominator:

{{{(3*log(2,3))/(3*log(2,2))}}}

Cancel the {{{3}}}'s:

{{{(cross(3)*log(2,3))/(cross(3)*log(2,2))}}}

and we have just:

{{{(log(2,3))/(log(2,2))}}}

Use the rule {{{log(B,(B))=1}}} to rewrite the denominator:

{{{(log(2,3))/1}}}

or just

{{{log(2,3)}}}

Now let's go back to the original problem, which was:

{{{4^(log(8,27))}}}

and replace the exponent with {{{log(2,3)}}}

{{{4^log(2,3)}}}

Rewrite the base {{{4}}} as {{{(2^2)}}}

{{{(2^2)^log(2,3)}}}

Use the rule {{{(B^M)^N=B^(M*N)}}}

{{{2^(2*log(2,3))}}}

Use the rule {{{N*log((B),X)=log((B),X^N)}}} to rewrite the exponent:

{{{2^(log(2,3^2))}}}

Write {{{3^2}}} as {{{9}}}

{{{2^(log(2,9))}}}

Use the rule {{{log(B,(B^N)) = N}}}

Final answer: 

{{{9}}}

--------------------------------

{{{drawing(200,100,0,2,0,-2,

locate(0,-1,4^((log(27)/log(8))))   ) }}}

Again we notice there is a {{{4}}} and an {{{8}}} involved,
and both are powers of {{{2}}}, let's try to get as much
as possible involving the number {{{2}}}:

As before we first work only with the exponent:

{{{log(27)/log(8)}}}

We write {{{27}}} as {{{3^3}}} and {{{8}}} as {{{2^3}}}

{{{log(3^3)/log(2^3)}}}

Use the rule {{{log((B),X^N)=N*log((B),X)}}} to rewrite
numerator and denominator:

{{{(3*log(3))/(3*log(2))}}}

Cancel the {{{s}}}'s

{{{(cross(3)*log(3))/(cross(3)*log(2))}}}

{{{log(3)/log(2)}}}

When no base is written, the base is assumed to be {{{10}}}

{{{log(10,3)/log(10,2)}}}

Write that fraction as a division:

{{{log(10,3)}}}{{{"÷"}}}{{{log(10,2)}}}

Use the change of base formula: {{{log((OLD_BASE),(X))=log((NEW_BASE),(X))/log((NEW_BASE),(OLD_BASE))}}} on each to get them
to logs base 2.

{{{log(2,3)/log(2,10)}}}{{{"÷"}}}{{{log(2,2)/log(2,10)}}}

Use the rule {{{log(B,(B)) = 1}}} to rewrite {{{log(2,2)}}}

{{{log(2,3)/log(2,10)}}}{{{"÷"}}}{{{1/log(2,10)}}}

Invert the second fraction and change division to multiplication:

{{{log(2,3)/log(2,10)}}}{{{"×"}}}{{{log(2,10)/1}}}

Cancel:

{{{log(2,3)/cross(log(2,10))}}}{{{"×"}}}{{{cross(log(2,10))/1}}}

All that's left is 

{{{log(2,3)}}}

Now let's go back to the original problem, which was:

{{{drawing(200,100,0,2,0,-2,

locate(0,-1,4^((log(27)/log(8))))   ) }}}

and replace the exponent by the simpler {{{log(2,3)}}}

{{{4^log(2,3)}}} 

Wow!  That just happens by chance to be the exact same
thing as we got in the first problem, so we just copy
the above from there on and, of course, it will have
the same final answer.

Rewrite the base {{{4}}} as {{{(2^2)}}}

{{{(2^2)^log(2,3)}}}

Use the rule {{{(B^M)^N=B^(M*N)}}}

{{{2^(2*log(2,3))}}}

Use the rule {{{N*log((B),X)=log((B),X^N)}}} to rewrite the exponent:

{{{2^(log(2,3^2))}}}

Write {{{3^2}}} as {{{9}}}

{{{2^(log(2,9))}}}

Use the rule {{{log(B,(B^N)) = N}}}

Final answer: 

{{{9}}} 

Edwin</pre>