Question 842219
{{{log(2,(35))}}}{{{""-""}}}{{{log(4,(13))}}}

Use change of base formula on second term:

{{{log(A,(C))=log(B,(C))/log(B,(A))}}}

{{{log(2,(35))}}}{{{""-""}}}{{{log(2,(13))/log(2,(4))}}}

We calculate {{{log(2,(4))}}} by setting it equal to x

{{{log(2,(4))}}}{{{""=""}}}{{{x}}}

by using the fact that {{{log(B,(A))=C}}} is equivalent to {{{B^C=A}}}

{{{2^x}}}{{{""=""}}}{{{4}}}
{{{2^x}}}{{{""=""}}}{{{2^2}}}
Equate exponents
x = 2

So {{{log(2,(4))}}}{{{""=""}}}{{{2}}} 

{{{log(2,(35))}}}{{{""-""}}}{{{log(2,(13))/2}}}

{{{log(2,(35))}}}{{{""-""}}}{{{expr(1/2)*log(2,(13))}}}

Use the rule {{{A*log(B,(C))=log(B,(C^A))}}}
 
{{{matrix(2,1,"",log(2,(35)))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",


log(2,(13^(1/2))))}}}

The {{{1/2}}} power is the same as the square root:

{{{log(2,(35))}}}{{{""-""}}}{{{log(2,(sqrt(13)))}}}

Use the rule {{{log(B,(A))-log(B,(C))=log(B,(A/C))}}}

{{{log(2,(35/sqrt(13)))}}}

Edwin</pre>