Question

How do you show `log_(ab)p=(mn)/(m+n)` ?

Given that `log_ap=m` and `log_bp=n`.

Answer 1

Please see below.

Explanation:

Given `log_ap=m` and `log_bp=n`, we have `a^m=p` and `b^n=p`. Hence

`(ab)^(mn)=a^(mn)b^(mn)`

= `p^n*p^m`

= `p^(m+n)`

or `(ab)^((mn)/(m+n))=p`

Hence `log_(ab)p=(mn)/(m+n)`

Answer 2

Please see below.
Using Law of logarithms

The shortest method , using Exponent Rules is given by respected @Shwetank Mauria for this answer.

Explanation:

We know that

`color(red)((1)log_xy=log_ky/log_kx,where, k` .is new base`to["Change of Base"]`

`color(blue)((2)log_z(x*y)=log_zx+log_zy..to["Product/sum rule]"`

Given that,

`color(red)(log_ap)=m and color(red)(log_bp)=n`

Using `(1)` we get

`color(red)(log_kp/log_ka)=m and color(red)(log_kp/log_kb)=n`

`=>1/m=log_ka/log_kp and 1/n=log_kb/log_kp`

So,

`1/m+1/n=log_ka/log_kp+log_kb/log_kp`

Simplifying we get,

`(n+m)/(mn)=(log_ka+log_kb)/log_kp`

`=>(m+n)/(mn)=log_k(ab)/log_kp...tocolor(blue)(Apply(2)`

`=>(mn)/(m+n)=log_kp/log_k(ab)`

Again using `(1)` to RHS ,we get

`(mn)/(m+n)=log_(ab)p`

`i.e. log_(ab)p=(mn)/(m+n)`

Note:

Change of Base formula:

`color(red)(log_(ab)p=log_kp/log_k(ab)`