Solve 7^2x-1=7^x+5?

Question

and also solve (1/32)^x-8=16^x

Answer 1

since they have the same base which is 7.. you could equate the exponents to each other

2x-1 =x+5
x=6



(1/32)^(x-8) = 16 ^x ..... make the base the same..

(1/2^(5))^ (x-8) =(2^4)^x <-- so the base now is 2

(2^-5)^(x-8) =(2^4)^x <-- to simplify.. multiply the exponents

2^(-5x+40) =2^(4x) <-- since the base is the same (2).. you could equate the exponent to each other

-5x +40 =4x
40=9x
40/9=x

Answer 2

Solve 7^2x-1=7^x+5?put 7^x =y, we obtain:
y^2 -1 =y+5
or y^2 -y -6 =0
y=3,-2
7^x =3 x=log3 to the base 7.

Answer 3

7^(2x-1) = y^(x+5)
equal bases, equal values, must be equal exponents:

2x-1 = x+5
x = 6

(1/32)^(x-8) = 16^x
(2^-5)^(x-8) = (2^4)^x
2^(-5x+40) = 2^(4x)
-5x + 40 = 4x
40 = 9x
x = 40/9

when you have expressions for exponents, you really ought to use () around them.

Answer 4

x = 6
x = 40/9