Question 202615
{{{2/log(4,(2000^6))+3/log(5,(2000^6))}}} Start with the given expression.



Let {{{x=2000^6}}}



{{{2/log(4,(x))+3/log(5,(x))}}} Replace every instance of {{{2000^6}}} with "x". This isn't necessary, but I find this substitution simplifies things a bit.



{{{2/(log(10,(x))/log(10,(4)))+3/(log(10,(x))/log(10,(5)))}}} Use the change of base formula.



Note: the change of base formula is {{{log(b,(x))=log(10,(x))/log(10,(b))}}}
 



{{{2*(log(10,(4))/log(10,(x)))+3*(log(10,(5))/log(10,(x)))}}} Multiply by the reciprocal of the lower fractions.



{{{(2*log(10,(4)))/log(10,(x))+(3*log(10,(5)))/log(10,(x))}}} Multiply.



{{{(2*log(10,(4))+3*log(10,(5)))/log(10,(x))}}} Combine the fractions.



{{{(log(10,(4^2))+log(10,(5^3)))/log(10,(x))}}} Rewrite each log using the identity  {{{y*log(b,(x))=log(b,(x^y))}}}



{{{(log(10,(16))+log(10,(125)))/log(10,(x))}}} Square 4 to get 16 and cube 5 to get 125. note: {{{5^3=125}}} (not 625)



{{{(log(10,(16*125)))/log(10,(x))}}} Combine the logs using the identity {{{log(b,(A))+log(b,(B))=log(b,(A*B))}}}



{{{(log(10,(2000)))/log(10,(x))}}} Multiply



{{{(log(10,(2000)))/log(10,(2000^6))}}} Plug in {{{x=2000^6}}}



{{{(log(10,(2000)))/(6*log(10,(2000)))}}} Rewrite the bottom log using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{(highlight(log(10,(2000))))/(6*highlight(log(10,(2000))))}}} Highlight the common terms.



{{{(cross(log(10,(2000))))/(6*cross(log(10,(2000))))}}} Cancel out the common terms.



{{{1/6}}} Simplify



So {{{2/log(4,(2000^6))+3/log(5,(2000^6))=1/6}}}



You can verify this with a calculator.