Solve for x. log(x)32 = 5?

Question

i changed it to

x^5 = 32 but dont know how to reduce if further....

same with log(2)64= x

i got 2^x = 64 but didnt know what to do next...

Answer 1

For x^5 = 32, just take the fifth root of both sides
(x^5)^(1/5) = 32^(1/5)
x = 32^(1/5)
x = (4*8)^(1/5)
x = [2^2 2^3]^(1/5)
x = [2^5]^(1/5)
x = 2


For log[base 2](64), I don't know why you're putting it back into exponents. It's already solved for you. But you can simplify it from there by getting 64 as a power of 2:
2^x = 64
2^x = 8^2
2^x = (2^3)^2
2^x = 2^6
So x = 6

Answer 2

To put it in layman terms: once you get x^5=32, take the fifth root of both sides to single out x ==> x = 32^ (1/5) = 2.

As you can see, 2^5 = 32. Based on your second problem, 64 is twice of 32, so if you multiply both sides of 2^5 =32 by 2, you get 2^6 = 64.

Answer 3

2^5 = 32, so x = 2

2^x = 64; 2^6 = 64, so x = 6

Answer 4

x=5/32

Answer 5

log(x)32 = 5

change to exponient form
x^5 = 32

take a fifth root
x = 2


2^x = 64

take log for both sides
log 2^x = log64

power property
x log2 = log64

x = log64 / log2

Answer 6

you can reduce 32 to 16x2 or 8x2x2 or 4x2x2x2 or 2x2x2x2x2
that makes 2^5 = x^5
So x=2

For log(2)64=x
2^x = 64
2^x = 2^6
x=6

Answer 7

Two raised to the fifth power is thirty two. I suspect that's what you're looking for.