Math, asked by Rahul11105, 2 months ago

The sum of the possible values of x of the equation
4^x - 3 [ 2^(x + 2) ] + 32 = 0 is
(1) 12
(2) 4
(3) 5
(4) 1​

Answers

Answered by tennetiraj86
2

Answer:

Option 3 i.e. 5

Step-by-step explanation:

Given:-

equation 4^x - 3 [ 2^(x + 2) ] + 32 = 0

To find:-

The sum of the possible values of x of the equation 4^x - 3 [ 2^(x + 2) ] + 32 = 0

Solution:-

Given equation is 4^x - 3 [ 2^(x + 2) ] + 32 = 0

=>(2^2)^x - 3[2^(x+2)] +32 = 0

we know that (a^m)^n = a^mn

=>2^2x -3[2^(x+2)] +32 = 0

We know that a^(m+n)=a^m × a^n

=>2^2x - 3[2^x × 2^2] + 32 = 0

=>2^2x -3[2^x ×4] +32 = 0

=>2^2x -12×2^x +32 = 0

=>(2^x)^2 - 12×2^x +32 = 0

Put 2^x = a then it becomes

=>a^2 -12a +32 = 0

=>a^2 -4a -8a +32 = 0

=>a(a-4)-8(a-4) = 0

=>(a - 4)(a - 8) = 0

=>a - 4 = 0 (or) a - 8 = 0

=>a = 4 or a = 8

now

=>2^x = 4 or 2^x = 8

=>2^x = 2^2 or 2^x = 2^3

Since bases are equal then exponents must be equal

=>x = 2 and x = 3

The possible values of x = 2 and 3

The sum of the values of x = 2+3 = 5

Answer:-

The sum of the possible values of x = 5

Used formula:-

  • a^m × a^n = a^(m+n)
  • (a^m)^n = a^mn

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