Question

$\tt: \red{ {25}^{n - 1} + 100 = {5}^{2n - 1}}$ Find value of n

✊Don't spam! ​

Answer 1

Step-by-step explanation:

25^n-1 + 100 = 5^2n-1

25^n.25^-1 + 100= 5^2n.5^-1

25^n.25^-1 + 100= 25^n.5^-1

Now take 25^n.25^-1 part to RHS then will get

25^n.5^-1 - (25^n.25^-1 )=100

Now take 25^n common from LHS

25^n(1/5 - 1/25)= 100

25^n(4/25)=100

25^n=100*25/4

25^n=25*25

25^n=25^2

Now we can say that

n=2.

Answer 2

Given :-

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

To Find :-

• $\sf n = ?$

Solution :-

$\qquad\leadsto\quad\sf \pink{ {25}^{n - 1} + 100 = {5}^{2n - 1}}$

$\qquad\leadsto\quad\sf100 = {5}^{2n - 1} - {25}^{n - 1}$

$\qquad\leadsto\quad\sf {5}^{2n - 1} - {25}^{n - 1} = 100$

$\qquad\leadsto\quad\sf {5}^{2n} \times {5}^{ - 1} - {25}^{n} \times {25}^{ - 1} = 100$

$\qquad\leadsto\quad\sf({5}^{2} ) ^{n} \times \frac{1}{5} - {25}^{n} \times \frac{1}{25} = 100$

$\qquad\leadsto\quad\sf \dfrac{ {25}^{n} }{5} - \frac{ {25}^{n} }{25} = 100$

$\qquad\leadsto\quad\sf \dfrac{5 \times {25}^{n} - {25}^{n} }{25} = 100$

$\qquad\leadsto\quad\sf {25}^{n} (5 - 1) = 100 \times 25$

$\qquad\leadsto\quad\sf {25}^{n} \times 4 =2500$

$\qquad\leadsto\quad\sf {25}^{n} = \frac{2500}{4}$

$\qquad\leadsto\quad\sf{25}^{n} = 625$

$\qquad\leadsto\quad\sf{25}^{n} = {25}^{2}$

$\qquad\leadsto\quad\sf \pink{n = 2}$

$\therefore\:\underline{\textsf{Value of n is \textbf{2}}}.$