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

3 Answers
Jumbotron
Feb 22, 2018

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!

Jim G.
Feb 22, 2018

"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

Surya K.
Feb 22, 2018

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.

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