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

Feb 22, 2018

see a solution process below;

#### Explanation:

${3}^{m} = 2 \mathmr{and} {4}^{n} = 27$

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

${3}^{m} = 2$

Log both sides..

$\log {3}^{m} = \log 2$

$m \log 3 = \log 2$

$m = \log \frac{2}{\log} 3$

similarly..

${4}^{n} = 27$

Log both sides..

$\log {4}^{n} = \log 27$

$n \log 4 = \log 27$

$n = \log \frac{27}{\log} 4$

$n = \log {3}^{3} / \log {2}^{2}$

$n = \frac{3 \log 3}{2 \log 2}$

Hence;

$m \times n \Rightarrow \log \frac{2}{\log} 3 \times \frac{3 \log 3}{2 \log 2}$

$m \times n \Rightarrow \cancel{\log} \frac{2}{\cancel{\log}} 3 \times \frac{3 \cancel{\log} 3}{2 \cancel{\log} 2}$

$m \times n \Rightarrow \frac{3}{2}$

As required!

Feb 22, 2018

$\text{see explanation}$

#### Explanation:

${4}^{n} = {\left({2}^{2}\right)}^{n} = {\left(2\right)}^{2 n} = 27 = {3}^{3} \leftarrow \text{from } {4}^{n} = 27$

$\text{substitute } 2 = {3}^{m}$

$\Rightarrow {\left({3}^{m}\right)}^{2 n} = {3}^{3}$

$\Rightarrow {3}^{2 m n} = {3}^{3}$

$\text{since bases on both sides are 3, equate the exponents}$

$\Rightarrow 2 m n = 3$

$\Rightarrow m n = \frac{3}{2}$

Feb 22, 2018

See below.

#### Explanation:

We have ${4}^{n} = 27$

We can write that as:

${\left({2}^{2}\right)}^{n} = {3}^{3}$

${2}^{2 n} = {3}^{3}$

Since ${3}^{m} = 2$, we can input:

${\left({3}^{m}\right)}^{2 n} = {3}^{3}$

${3}^{2 m n} = {3}^{3}$

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

$2 m n = 3$

$m n = \frac{3}{2}$

Proved.