Question

If `3^m = 2 and 4^n =27`, show by laws of indices that `m xx n =3/2`??

Answer 1

see a solution process below;

Explanation:

`3^m = 2 and 4^n = 27`

Note, we can also use Law or Logarithm to solve this;

`3^m = 2`

Log both sides..

`log3^m = log2`

`mlog3 = log2`

`m = log2/log3`

similarly..

`4^n = 27`

Log both sides..

`log4^n = log27`

`nlog4 = log27`

`n = log27/log4`

`n = log3^3/log2^2`

`n = (3log3)/(2log2)`

Hence;

`m xx n rArr log2/log3 xx (3log3)/(2log2)`

`m xx n rArr cancellog2/cancellog3 xx (3cancellog3)/(2cancellog2)`

`m xx n rArr 3/2`

As required!

Answer 2

`"see explanation"`

Explanation:

`4^n=(2^2)^n=(2)^(2n)=27=3^3larr"from "4^n=27`

`"substitute "2=3^m`

`rArr(3^m)^(2n)=3^3`

`rArr3^(2mn)=3^3`

`"since bases on both sides are 3, equate the exponents"`

`rArr2mn=3`

`rArrmn=3/2`

Answer 3

See below.

Explanation:

We have `4^n=27`

We can write that as:

`(2^2)^n=3^3`

`2^(2n)=3^3`

Since `3^m=2`, we can input:

`(3^m)^(2n)=3^3`

`3^(2mn)=3^3`

Since the bases are equal, the exponents are equal too.

`2mn=3`

`mn=3/2`

Proved.