Question

if
${25}^{n - 1} + 100 = {5}^{2n - 1}$
find the value of n​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

$\sf{25^{(n-1)} + 100 = 5^{(2n - 1)}}$

$\sf{(5)^{2(n - 1)} + 100 = 5^{(2n - 1)}}$

$\sf{5^{(2n - 2)} + 100 = 5^{(2n - 1)}}$

$\sf{5^{(2n - 1)} - 5^{(2n - 2)} = 100}$

$\sf{5^{(2n - 1)}(1 - 5^{-1}) = 100}$

Using Exponential :-

$\sf{(a^{-n}) =(\dfrac{1}{a^{n}})}$

$\sf{Now \: ,}$

$\sf{5^{(2n - 1)}(1 - \dfrac{1}{5}) = 100}$

$\sf{5^{(2n - 1)}(\dfrac{4}{5}) = 100}$

$\sf{5^{(2n - 1)} = 100\times \dfrac{5}{4}}$

$\sf{5^{(2n - 1} = 25\times 5}$

$\sf{5^{(2n - 1)} = 125}$

$\sf{5^{(2n - 1)} = 5^{3}}$

$\sf{So \: ,}$

$\sf{(2n - 1) = 3}$

$\sf{2n = 4}$

$\sf{n = 2}$

Extra important Exponential :-

1)$\sf{a^{m}\times a^{n}=a^{m+n}}$

2)$\sf{\dfrac{a^{m}}{a^{n}} = a^{m-n}}$

3)$\sf{(a^{m})^{n} = a^{mn}}$

4)$\sf{a^{m}\times b^{m} = (ab)^{m}}$

5)$\sf{a^{0} = 1}$