Question

Q7. If 2ˣ⁻¹ + 2ˣ⁺¹ = 320, then find x.​

Answer 1

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Answer 2

Step-by-step explanation:

Given :-

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

To find :-

Find the value of x ?

Solution:-

Given equation is 2^(x-1) +2^(x+1) = 320

We know that

a^m×a^n = a^(m+n)

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

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

=>(2^x/2^1) +(2^x×2^1) = 320

=> (2^x/2 ) +(2×2^x) = 320

=> (2^x+4×2^x)/2 = 320

=> 2^x + 4 × 2^x = 320×2

=> 2^x + 4 × 2^x = 640

=>>(1+4)×2^x = 640

=> 5×2^x = 640

=> 2^x = 640/5

=> 2^x = 128

=> 2^x = 2×2×2×2×2×2×2

=> 2^x = 2^7

If bases are equal then exponents must be equal.

x = 7

Answer:-

The value of x for the given problem is 7

Check:-

If x = 7 then LHS in the given equation

=> 2^(x-1) +2^(x+1)

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

=> 2^6 + 2^8

=> 64 + 256

=> 320

LHS = RHS is true for x = 7

Verified the given relation in the given problem.

Used formulae:-

• a^m×a^n = a^(m+n)
• a^m/a^n = a^(m-n)