Question 1206690
<br>
Ouch!!!<br>
The solution method shown by the other tutor is valid, but WAY more work than needed....<br>
{{{log(2,(x))+log(4,(x))+log(16,(x))=21/4}}}<br>
The bases of the logarithms are all powers of 2, making for a quick and easy path to the answer.<br>
{{{16=2^4}}}, so {{{log(2,(x))=4*log(16,(x))}}}
{{{16=4^2}}}, so {{{log(4,(x))=2*log(16,(x))}}}<br>
So<br>
{{{log(2,(x))+log(4,(x))+log(16,(x))=4*log(16,(x))+2*log(16,(x))+log(16,(x))=7*log(16,(x))}}}<br>
Then<br>
{{{7*log(16,(x))=21/4}}}
{{{log(16,(x))=3/4}}}
{{{x=16^(3/4)=8}}}<br>
ANSWER: 8<br>