Question 254894
Use several change of bases:
(i) log_A(x) = C
By change of base rules, we get
(ii) log(x)/log(A) = C
and solving for log(A) we get
(iii) log(A) = log(x)/C
--
(iv) log_B(x) = D
By change of base rules, we get
(v) log(x)/log(B) = D
and solving for log(B) we get
(vi) log(B) = log(x)/D
--
Now,
(vii) log_AB(x) 
becomes
(viii) log(x)/log(AB) 
which is expanded to
(ix) log(x) / (log(A) + log(B)).
So,
by substitution of (iii) and (vi) into (ix), we get
(x) {{{log(x) / (log(x)/C + log(x)/D)}}}
by adding the denominator fractions, we get
(xi){{{log(x) / ((Dlog(x) + Clog(x))/CD)}}}
factoring out the log(x) gives us
(xii) {{{log(x) / (log(x)(C + D))div(CD)}}}
and finally the answer as
{{{CD/(C+D))}}}