Find the values of x that makes the equation

Question

49^x+1 = 7^3x^2-6 true.?

Answer 1

Given equation : 49^x+1 = 7^3x²-6

(7²)^x+1 = 7^3x²-6

7^2(x+1) = 7^3x²-6

Bases equal then exponents also equal.

2(x+1) = 3x² - 6

2x + 2 = 3x² - 6

3x² – 6 – 2x – 2 = 0

3x² – 2x – 8 = 0

3x² – 6x + 4x – 8 = 0

3x(x – 2) + 4(x – 2) = 0

(x – 2) (3x +4) = 0

By using zero product property : If AB = 0 then A = 0 , B = 0

(x – 2) = 0  and   (3x +4) = 0

x = 2 , - 4/3

Solution is x = 2 , - 4/3