EXPONENTIAL EQUATION

Question

(1/7)^X-1=49X

Answer 1

Is 49x supposed to be 49^x?

Also, is (1/7)^x-1 supposed to be (1/7)^(x-1)?

 

If so, here is a solution:

(1/7)^{(x-1) =} 49^x

7^(-1)(x-1) = 7^2x

7^(-x+1) = 7^2x      Since the bases are equal, the exponents are equal.

-x+1 = 2x

1 = 3x

1/3 = x