Question

Find the value of 'n' if: $(i) \: {3}^{n - 5} × {6}^{n - 3} = 36 \\ (ii) \: {2}^{n - 4} × {4}^{n - 2} = 16 \\ (iii) \: {5}^{n - 1} × {25}^{n - 2} = 125$​

Answer 1

$\large\underline{\sf{Solution-(i)}}$

Given expression is

$\rm \: {3}^{n - 5} × {6}^{n - 3} = 36$

can be rewritten as

$\rm \: {3}^{n - 5} × {(3 \times 2)}^{n - 3} = 3 \times 3 \times 2 \times 2$

$\rm \: {3}^{n - 5} × {3}^{n - 3} \times {2}^{n - 3} = {3}^{2} \times {2}^{2}$

We know,

$\boxed{ \rm{ \: {x}^{m} \: \times \: {x}^{n} \: = \: {x}^{m + n} \: \: }}$

So, using this result, we get

$\rm \: {3}^{n - 5 + n - 3} \times {2}^{n - 3} = {3}^{2} \times {2}^{2}$

$\rm \: {3}^{2n - 8} \times {2}^{n - 3} = {3}^{2} \times {2}^{2}$

So, on comparing, we get

$\rm \: n - 3 = 2$

$\rm\implies \:\boxed{ \rm{ \:n \: = \: 5 \: }}$

$\large\underline{\sf{Solution-(ii)}}$

Given expression is

$\rm \: {2}^{n - 4} × {4}^{n - 2} = 16$

can be rewritten as

$\rm \: {2}^{n - 4} × {( {2}^{2} )}^{n - 2} = {2}^{4}$

$\rm \: {2}^{n - 4} × {2}^{2n - 4} = {2}^{4}$

We know,

$\boxed{ \rm{ \: {x}^{m} \: \times \: {x}^{n} \: = \: {x}^{m + n} \: \: }}$

So, using this identity, we get

$\rm \: {2}^{n - 4 + 2n - 4} = {2}^{4}$

$\rm \: {2}^{3n - 8} = {2}^{4}$

So, on comparing, we get

$\rm \: 3n - 8 = 4$

$\rm \: 3n = 4 + 8$

$\rm \: 3n = 12$

$\rm\implies \:\boxed{ \rm{ \:n \: = \: 4 \: }}$

$\large\underline{\sf{Solution-(iii)}}$

Given expression is

$\rm \: {5}^{n - 1} × {25}^{n - 2} = 125$

can be rewritten as

$\rm \: {5}^{n - 1} × {( {5}^{2}) }^{n - 2} = {5}^{3}$

$\rm \: {5}^{n - 1} × {5}^{2n - 4} = {5}^{3}$

We know,

$\boxed{ \rm{ \: {x}^{m} \: \times \: {x}^{n} \: = \: {x}^{m + n} \: \: }}$

So, using this identity, we get

$\rm \: {5}^{n - 1 + 2n - 4} = {5}^{3}$

$\rm \: {5}^{3n - 5} = {5}^{3}$

So, on comparing we get

$\rm \: 3n - 5 = 3$

$\rm \: 3n = 3 + 5$

$\rm \: 3n = 8$

$\rm\implies \:\boxed{ \rm{ \:n \: = \: \frac{8}{3} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Question (i):-

$\text{Given: \( \rm 3^{n-5} \times 6^{n-3}=36 \)}$

To find: The value of n

Solution:-

$\[ \begin{array}{rlr} & \rm 3^{n-5} \times 6^{n-3}=36 & \\ \\ \Rightarrow & \rm 3^{n-5} \times(3 \times 2)^{n-3}=36 & \\ \\ \Rightarrow & \rm 3^{n-5} \times 3^{n-3} \times 2^{n-3}=36 & \rm \left((a b)^{m}=a^{m} \times b^{m}\right) \\ \\ \Rightarrow & \rm 3^{n-3}+n-3 \times 2^{n-3}=36 & \rm \left(a^{m} \times a^{n}=a^{m+n}\right) \\ \\ \Rightarrow & \rm 3^{2 n-8} \times 2^{n-3}=36 & \end{array} \]$

The R.H.S. cannot be solved any futher. So, we Should break 36 into a product of primes and then comparing the powers we can find the value of n .

$\[ \begin{array}{l} \rm36=2^{2} \times 3^{2} \\ \\ \rm \Rightarrow 3^{2 n-8} \times 2^{n-3}=3^{2} \times 2^{2} \end{array} \]$

Comparing L.H.S. and R.H.S., we get two equations:-

$\[ \begin{aligned} \rm 2 n-8 & \rm=2 \quad \text { and } \quad n-3=2 \\ \\ \rm 2 n & \rm=10 \quad \text { and } \quad n=5 \\ \\ \rm \Rightarrow \quad n &=5 \end{aligned} \]$

Therefore, the value of n from both equations is 5 . Hence, the value of n=5.

Answer (ii):-

$\\ \rm\implies {2}^{n - 4} \times {4}^{n - 2}$

Convert both sides to the same base.

$\\ \rm\implies {2}^{n - 4} \times {2}^{2( n- 2)} = {2}^{4}$

Use Product Rule: $\rm{x}^{a}{x}^{b}={x}^{a+b}$

$\\ \rm\implies {2}^{n-4+2(n-2)}={2}^{4}$

$\\ \rm\implies \: n−4+2(n−2)=4$

$\\ \rm\implies n−4+2n−4=4$

$\\ \rm\implies 3n−8=4$

$\\ \rm\implies3n=4+8$

$\\ \rm\implies 3n = 12$

$\\ \rm\implies n=\frac{12}{3}$

$\\ \boxed{\red{ \rm\implies \: n = 4}}$

Answer (iii):-

$\[ \begin{array}{l} \rm 5^{n-1} \times 25^{n-2}=125 \\ \\ \rm \Rightarrow 5^{n-1} \times\left(5^{2}\right)^{n-2}=125 \\ \\ \rm\Rightarrow 5^{n-1} \times 5^{2(n-2)}=125 \\\\ \rm \Rightarrow 5^{n-1} \times 5^{2 n-4}=5^{3} \\ \\ \rm\Rightarrow 5^{(n-1)+(2 n-4)}=5^{3} \end{array} \]$

On comparing powers:

$\[ \begin{array}{l} \rm \Rightarrow(n-1)+(2 n-4)=3 \\ \\ \rm\Rightarrow n-1+2 n-4=3 \\ \\ \rm \Rightarrow 3 n-5=3 \\ \\ \rm\Rightarrow 3 n=3+5 \\ \\ \rm \Rightarrow 3 n=8 \\ \\ \boxed{ \color{orangered} \displaystyle \rm\Rightarrow n=\frac{8}{3} }\end{array} \]$

Hence, value of n is $\rm \dfrac{8}{3}.$