Question 222790
I found that it helps to work backwards for a bit to find out how to go forward. In this case, I broke down {{{log(ab,(x))}}} to get:


{{{log(ab,(x))}}} 



{{{log(10,(x))/log(10,(ab))}}} (used the change of base formula here)



{{{log(10,(x))/(log(10,(a))+log(10,(b)))}}} (used the identity {{{log(b,(A*B))=log(b,(A))+log(b,(B))}}})



Then I realized that I needed to find substitutions for {{{log(10,(a))}}} and {{{log(10,(b))}}} (to somehow get 'x' in there). So I figured I'd solve for those respective expressions like so:




{{{log(a,(x))=C}}} Start with the first equation.



{{{log(10,(x))/log(10,(a))=C}}} Use the change of base formula.



{{{log(10,(x))=C*log(10,(a))}}} Multiply both sides by {{{log(10,(a))}}}.



{{{log(10,(x))/C=log(10,(a))}}} Divide both sides by C.



So this means {{{log(10,(a))=log(10,(x))/C}}}



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{{{log(b,(x))=D}}} Start with the second equation.



{{{log(10,(x))/log(10,(b))=D}}} Use the change of base formula.



{{{log(10,(x))=D*log(10,(b))}}} Multiply both sides by {{{log(10,(b))}}}



{{{log(10,(x))/D=log(10,(b))}}} Divide both sides by D.



So this means {{{log(10,(b))=log(10,(x))/D}}}


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Now let's get back to the main problem:



{{{log(ab,(x))}}} Start with the given expression.



{{{log(10,(x))/log(10,(ab))}}} Use the change of base formula.



{{{log(10,(x))/(log(10,(a))+log(10,(b)))}}} Use the identity {{{log(b,(A*B))=log(b,(A))+log(b,(B))}}} to break up the denominator.



Now here's where the substitutions come into play:



{{{log(10,(x))/(log(10,(x))/C+log(10,(x))/D)}}} Plug in {{{log(10,(a))=log(10,(x))/C}}} and {{{log(10,(b))=log(10,(x))/D}}}



From here, it's just boils down to algebraically simplifying the expression:



{{{log(10,(x))/((D*log(10,(x)))/(CD)+log(10,(x))/D)}}} Multiply the first inner fraction by {{{D/D}}}.



{{{log(10,(x))/((D*log(10,(x)))/(CD)+(C*log(10,(x)))/(CD))}}} Multiply the second inner fraction by {{{C/C}}}.



{{{log(10,(x))/((D*log(10,(x))+C*log(10,(x)))/(CD))}}} Combine the lower fractions.



{{{log(10,(x))*((CD)/(D*log(10,(x))+C*log(10,(x))))}}} Multiply {{{log(10,(x))}}} by the reciprocal of the lower fraction.



{{{(CD*log(10,(x)))/(D*log(10,(x))+C*log(10,(x)))}}} Multiply



{{{(CD*log(10,(x)))/(log(10,(x))(D+C))}}} Factor out the GCF {{{log(10,(x))}}} from the denominator.



{{{(CD*cross((log(10,(x)))))/(cross((log(10,(x))))(D+C))}}} Cancel out the common terms.



{{{(CD)/(D+C)}}} Simplify



{{{(CD)/(C+D)}}} Rearrange the terms (trivial, but I figured that I'd match the book).



So this means {{{log(ab,(x))=(CD)/(C+D)}}} where {{{log(a,(x))=C}}} and {{{log(b,(x))=D}}}