Question 170951
Here's one way to do it:


{{{(log(7,(4)))(log(4,(9)))=log(7,(9))}}} Start with the given equation. Note: I'm ONLY going to manipulate the left side. The right side is for comparison



By the change of base formula {{{log(b,(x))=log(10,(x))/log(10,(b))}}}, we can rewrite {{{log(7,(4))}}} into {{{(log(10,(4)))/(log(10,(7)))}}}



Also, we can rewrite {{{log(4,(9))}}} into {{{(log(10,(9)))/(log(10,(4)))}}}



{{{((log(10,(4)))/(log(10,(7))))((log(10,(9)))/(log(10,(4))))=log(7,(9))}}} Use the change of base formula to expand the logarithms (see above)



{{{(highlight(log(10,(4)))/(log(10,(7))))((log(10,(9)))/highlight(log(10,(4))))=log(7,(9))}}} Highlight the common terms.



{{{(cross(log(10,(4)))/(log(10,(7))))((log(10,(9)))/cross(log(10,(4))))=log(7,(9))}}} Cancel out the common terms.



{{{(1/(log(10,(7))))((log(10,(7)))/1)=log(7,(9))}}} Simplify



{{{((log(10,(7)))/(log(10,(7))))=log(7,(9))}}} Combine and multiply the fractions.



{{{log(7,(9))=log(7,(9))}}} Use the change of base formula (in reverse) to simplify the left side




So this shows us that {{{(log(7,(4)))(log(4,(9)))=log(7,(9))}}}




Note: this can be generalized to say that 



{{{(log(x,(y)))(log(y,(z)))=log(x,(z))}}}