Question

[5^2x+1] =6×[5^x]-1 find value of x​

Answer 1

Answer:

hola mate

see the attachment

hope it helps u

Answer 2

Step-by-step explanation:

${5}^{2x + 1} = 6 \times {5}^{x} - 1 \\ {5}^{2x} \times 5 = 6 \times {5}^{x} - 1 \\ ( {5}^{x} ) {}^{2} \times 5 = 6 \times {5}^{x} - 1 \\ = let \: \: {5}^{x} = y \\ {y}^{2 } \times 5 = 6y - 1 \\ = 5 {y}^{2} - 6y + 1 = 0 \\ 5 {y}^{2 } - 5y - y + 1 = 0 \\ 5y(y - 1) - 1(y - 1) = 0 \\ (5y - 1)(y - 1) \\ \\ 5y - 1 = 0 \\ y = \frac{1}{5} \\ \\ y - 1 = 0 \\ y = 1$

5^x = 5^-1

x = -1

5^x = 5^0

x =1