Question

12. If 2^n-1 + 2^ n+1 = 320, then n is equal to (A) 6 (B) 8 (C) 5​

Answer 1

Step-by-step explanation:

Solution :-

Given that

2^(n-1) + 2^(n+1) = 320

=> (2^n/2) + (2^n×2) = 320

Since ,a^m × a^n = a^ (m+n)

and a^m / a^n = a^(m-n)

=> 2^n [ (1/2)+2 ] = 320

=> 2^n [ (1+4)/2 ] = 320

=> 2^n (5/2) = 320

=> 2^n = 320×2/5

=> 2^n = 640/5

=> 2^n = 128

=> 2^n = 2×2×2×2×2×2×2

=> 2^n = 2^7

We know that

If bases are equal then exponents must be equal.

Therefore, n = 7

Answer :-

The value of n is 7

Check:-

If n = 7 them LHS of the given equation then

2^(7-1)+ 2^(7+1)

= 2^6 + 2^8

= 64+256

= 320

= RHS

LHS = RHS is true for n = 7

Used formulae:-

→ a^m × a^n = a^ (m+n)

→ a^m / a^n = a^(m-n)

→ If a^m = a^n => m = n

Answer 2

Answer:

7

Step-by-step explanation:

$\huge \red{solution}$

$\sf {2}^{n - 1} + {2}^{n + 1} = 320 \\ \sf \: = {2}^{n - 1} (1 + {2}^{2} ) = 320 \\ \sf \: = {2}^{n - 1} \times 5 = 320 \\ \sf \: = {2}^{n - 1} = \frac{320}{5} = 64 \: \\ \sf \: = (2) ^{n - 1} = (2)^{6} \\ \sf \: = n = 7$