Solve for x in this logarithmic equation?

Question

solve for x:

log 8 - (1/3) log x = log 2

please show your work in order to get a "best answer" rating

Answer 1

log 8 - (1/3) log x = log 2
Exponent rule of logs:
log(8) - log(x^(1/3)) = log(2)
Subtraction/division rule of logs:
log(8 / (x^(1/3))) = log(2)
Subtract log(2) from both sides:
log(8 / (x^(1/3))) - log(2) = 0
Subtraction/division rule of logs:
log([8 / (x^(1/3))] / 2) = 0
log(4 / (x^(1/3))) = 0
Definition of log (let b be the base):
b^0 = 4 / (x^(1/3))
1 = 4 / (x^(1/3))
Multiply both sides by x^(1/3):
x^(1/3) = 4
Cube both sides:
x = 4^3
x = 64


Check answer:
log(8) - (1/3)log(64) = log(2)
log(8) - log(64^(1/3)) = log(2)
log(8) - log(4) = log(2)
log(8/4) = log(2)
log(2) = log(2)
Answer checks.

Answer 2

log 8 X = -3

Answer 3

hmmm..
let's try:

so:
transposing:
log 8 - log 2 = 1/3 log x
3(log 8 - log 2) = log x

getting the inverse:
log^-1 [3(log8 - log2)] = x
using the calculator:
x = 64

Answer 4

log 8 - (1/3) log x = log 2
log 8 + log x ^(-1/3) = log 2
log [8 x^(-1/3) ] = log 2
8 x^(-1/3) = 2
8 = 2 x^(1/3)
x^(1/3) = 4
x = 64

Answer 5

log8-(1/3)logx=log2=>
log8-log2=log(x^(1/3))=>
but loga-logb=log(a/b). So
log8-log2=log4
log4=log(x^(1/3))
x^(1/3)=4=>x=64.