Question 143986
Use the rule for base conversion:  {{{log(n,x)=(log(b,x))/(log(b,n))}}}


So:  {{{log(16,x)=(log(2,x))/(log(2,16))}}}, but since we know that if {{{log(b,x)=y}}} then {{{b^y=x}}}, so {{{log(2,16)=y}}} means that {{{2^y=16}}}.  Solving for {{{y}}} gives us {{{y=4}}}, so {{{log(2,16)=4}}}.  That means that {{{log(16,x)=(log(2,x))/4}}}.


Using a similar process, we get:


{{{log(8,x)=(log(2,x))/3}}} and {{{log(4,x)=(log(2,x))/2}}}


Now we can write:
{{{((log(2,x))/4)+((log(2,x))/3)+((log(2,x))/2)+log(2,x)=6}}}


Applying the common denominator of 12:
{{{(3(log(2,x))/12)+(4(log(2,x))/12)+(6(log(2,x))/12)+(12(log(2,x))/12)=6}}}


Simplifying:
{{{25*log(2,x)=72}}}
{{{log(2,x)=2.88}}}


Therefore {{{x=2^(2.88)}}}


That is the exact answer.  You can use the x^y function on your Windows calculator in scientific mode to calculate a numerical approximation.