If log4 5=a and log5 6=b, then log3 2 is equal to
a⋅b=log45⋅log56=log46=12log26=12(log23+log22)=12(log23+1)a\cdot b=\log_45\cdot \log_56=\log_46=\frac{1}{2}\log_26=\frac{1}{2}(\log_23+\log_22)=\frac{1}{2}(\log_23+1)a⋅b=log45⋅log56=log46=21log26=21(log23+log22)=21(log23+1)
2ab=log23+12ab=\log_23+12ab=log23+1
log23=2ab−1\log_23=2ab-1log23=2ab−1
log32=1log23=12ab−1\log_3 2=\frac{1}{\log_2 3}=\frac{1}{2ab-1}log32=log231=2ab−11
Answer: log32=12ab−1\log_32=\frac{1}{2ab-1}log32=2ab−11 .
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