Question

if 4^n-4^(n-1)=24 then find out the value of (2n)^n​

Answer 1

Answer:

${5}^{( \frac{5}{2} )}$

Step-by-step explanation:

${\underline{{\text{Given}}}}$

${4}^{(n)} - {4}^{(n - 1)} = 24$

${\underline{{\text{To Find:}}}}$

The value of ${2n}^{n}$

${\underline{{\text{Solution:}}}}$

${4}^{n} - {4}^{n - 1} = 24\\ \\ {4}^{n} - {4}^{n} . {4}^{ - 1} = 24 \\ \\ {4}^{n} (1 - {4}^{ - 1} ) = 24 \\ \\ {4}^{n} (1 - \frac{1}{4} ) = 24 \\ \\ {4}^{n} ( \frac{4 - 1}{4} ) = 24 \\ \\ {4}^{n} \times \frac{3}{4} = 24 \\ \\ {4}^{n} \times \frac{ \cancel3}{4} = \cancel{24} \\ \\ { ({2}^{2} )}^{n} = 4 \times 8 \\ \\ {2}^{2n} = 32 \\ \\ {2}^{2n} = {2}^{5} \\ \\ 2n = 5 \\ \\ n = \frac{5}{2}$

Now,

${(2n)}^{n} \\ \\ (2 \times \frac{5}{2})^{ \frac{5}{2} } \\ \\ {5}^{ (\frac{5}{2} )} \\ \\ \therefore \text{The required answer is 5 ^\frac{5}{2} }</p><p>$

${\underline{{\text{ Some Important Laws of Indices}}}}$

${a}^{n} . {a}^{m} = {a}^{(n + m)} \\ \\ {a}^{ - 1} = \frac{1}{a} \\ \\ \frac{{a}^{n}}{ {a}^{m} } = {a}^{(n - m)} \\ \\ {(a {}^{c} )}^{b} = {a}^{b \times c} = {a}^{bc} \\ \\ {a}^{ \frac{1}{x} } = \sqrt[x]{a} \\ \\ [\text{where all variables are real and greater than 0}]$