1.

Prove the following identities: (a) `(log_(a) n)/(log_(ab) n) = 1+ log_(a) b" "(b) log_(ab) x = (log_(a) x log_(b) x)/(log_(a) x + log_(b) x)`.

Answer» (a) `(log_(n) n)/(log_(ab) n) = (log_(n) ab)/(log_(n) a) = (log_(n) a+ log_(n) b)/(log_(n) a)`
` = 1+(log_(n) b)/(log_(n) a) = 1+log_(a) b`
(b) `(log_(a) xlog_(b) x)/(log_(a) x+log_(b) x) =(1/(log_(x) a)1/(log_(x) b))/(1/(log_(x) a)+1/(log_(x) b))=(1/(log_(x) a)1/(log_(x) b))/((log_(x) a + log_(x) b)/(log_(x) a log_(x) b))`
` = 1/(log_(x) a+log_(x) b)=1/(log_(x) ab)=log_(ab) x`


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