SOLVING EXPONENTIAL EQUATIONS WITH DIFFERENT BASES WTHOUT LOGARITHMS

Problem 1 :

Solve 

4^x - 3(2^x+2) + 2⁵ = 0

Solution :

4^x - 3(2^x+2) + 2⁵ = 0

(2²)^x - 3(2^x ⋅ 2²) + 32 = 0

(2^x)² - 12(2^x) + 32 = 0

Let t = 2^x

t² - 12t + 32 = 0

(t - 4)(t - 8) = 0

t = 4 and t = 8

Problem 2 :

Solve 

2^x-2 + 2^3-x = 3

Solution :

Problem 3 :

Solving 4^x ⋅ 2^y = 128 and 3^{3x + 2y} = 9^xy, we get the roots.

Solution :

4^x ⋅ 2^y = 128

(2²)^x⋅ 2^y = 2⁷

2^2x+y = 2⁷

2x + y = 7 -----(1)

y = 7 - 2x

3^{3x + 2y} = 3^2xy

3x + 2y = 2xy ---(2)

Applying the value of y in (2), we get

3x + 2(7 - 2x) = 2x(7 - 2x)

3x + 14 - 4x = 14x - 4x²

4x² - x - 14x + 14 = 0

4x² - 15x + 14 = 0

(x - 2)(4x - 7) = 0

x = 2 and x = 7/4

Problem 4 :

Solving 9^x = 3^y and 5 ^x+y+1 = 25^xy, we get the roots ?

Solution :

9^x = 3^y and 5 ^x+y+1 = 25^xy

Applying the value y from (1) in (2)

x + 2x + 1 = 2x(2x)

3x + 1 = 4x²

4x² - 3x - 1 = 0

(4x + 1) (x - 1) = .0

x = -1/4 and x = 1

Problem 5 :

for some value of a and b, then the value of x is

a)  8       b)  4       c)  6        d)  2

Solution :

Here the powers are equal, so we can equate the bases for some value of a and b. But we need to find the value of x.

By applying x = 8, on both sides we are not going to receive the same values.

On both sides of the equal sign, we will get the same value by applying x = 2. So, the answer is 2.

Problem 6 :

If 2^{x + y} = 2^{2x - y} = √8 then the respective values of x and y are ?

Solution :

2^{x + y} = 2^{2x - y} = (2^3)^1/2

2^{x + y} = 2^{2x - y} = 2^3/2

x + y = 2x - y 

x - 2x + y + y = 0

-x + 2y = 0------(1)

2x - y = 3/2 ------(2)

x = 2y

By applying the value of x in (2), we get

2(2y) - y = 3/2

3y = 3/2

y = 1/2

x = 2(1/2)

x = 1

Problem 7 :

Solution :