Question

4^x-3.2^x+8=0 then find x​

Answer 1

Step-by-step explanation:

Given:-

4^x-3.2^x+8=0

To find:-

Find the value of x?

Solution:-

Given equation is 4^x-3.2^x+8=0

=> (2^2)^x -3 . 2^x + 8 = 0

=> 2^2x - 3. 2^x +8 = 0

(Since (a^m)^n = a^mn)

=> (2^x)^2 -3.2^x +8 = 0

Put 2^x = a then

=> a^2-3a+8 = 0

=> a^2 - 2(a)(3/2) +8 = 0

=> a^2-2(a)(3/2) = -8

On adding (3/2)^2 both sides

=> a^2-2(a)(3/2)+(3/2)^2 = -8+(3/2)^2

=> [a - (3/2)]^2 = -8+(9/4)

=> [a - (3/2)]^2 = (-32+9)/4

=> [a - (3/2)]^2 = -23/4

=> a - (3/2) = ±√(-23/4)

=> a -(3/2) = ±√-23/2

=> a = (3/2)±√-23/2

=> a = (3±√-23)/2

=> a = (3+√-23)/2 and (3-√-23)/2

if a = (3+√-23)/2

=> 2^x = (3+√-23)/2

On taking logarithms both sides

=> log 2^x = log (3+√-23)/2

=> x log 2 = log (3+√-23) -log 2

=> x = [log (3+√-23) -log 2]/log 2

If a = (3-√-23)/2

2^x = (3+√-23)/2

On taking logarithms both sides

=> log 2^x = log (3-√-23)/2

=> x log 2 = log (3-√-23) -log 2

=> x = [log (3-√-23) -log 2]/log 2

Answer:-

x = [log (3+√-23) -log 2]/log 2 or

[log (3-√-23) -log 2]/log 2 for the given problem

Used formulae:-

• (a^m)^n = a^mn
• (a-b)^2=a^2-2ab+b^2
• log a - log b = log (a/b)

Used Method:-

• Completing the square method