Help 64^2x = 16 ^x+1?

Question

I've tried doing this problem but i'm having some problems however, I know the answer is a fraction. Please show me how to solve it.

64^2x = 16 ^x+1

Answer 1

just turn the 64 into 2, and turn 16 into 2. you know 2^6 = 64, and you know 2^4 = 16. Now we have (2^6)^2x and (2^4)^x+1

Now multiply the exponents to get 2^12x and 2^ 4x + 4

now just do 12x = 4x + 4

Solve thsi to give 8x = 4
x = 1/2 or 0.5

Answer 2

I'm assuming "x+1" is all in the exponent. It's a far more difficult problem otherwise.

64^2x = 16^(x+1)
(2^6)^2x = (2^4)^(x+1)
2^(6*2x) = 2^(4*(x+1))
2^(12x) = 2^(4x+4)
12x = 4x + 4
8x = 4
x = 1/2

Answer 3

64^(2x) = 16^(x + 1)
Rewrite 64 as 2^6 and 16 as 2^4
[2^6]^(2x) = [2^4]^(x + 1)
Power to a power, multiply exponents
2^(12x) = 2^(4x + 4)
Exponential function is one to one so
if a^m = a^n, then m = n. Drop bases
12x = 4x + 4
8x = 4
x = 1/2

Answer 4

64^2x = 16^x+1
64^2x = 4^2x + 1
64^2x - 4^2x = 1

64^2x - 4^2x = 1
(64^2x)(4^2x) = 1
16^2x = 1
2x log 16 = 1
2x = 1/log 16
x = 1/(2 log 16)

Answer 5

Can you repeat that in plain English please?

This guy's a genius but he's giving me a headache, lol!

Looks like the answer is... 1/2 or 0.5!!!

Yeah right!

Answer 6

(4^3)^2x=4^2^x+1
4^6x=4^2x+1
6x=2x+1
4x=1
x=1/4