How to solve log8/log2? working out too plz?

Identity used:
log(aⁿ) = n log(a)

log(8) / log(2)
= log(2³) / log(2)
= 3 log(2) / log(2)
= 3

[Note that log(8)/log(2) ≠ log(8/2)]

= Log(8) / Log(2) → you know that: 8 = 2 * 2 * 2 = 2^3

= Log(2^3) / Log(2) → you know that: Log(x^a) = a.Log(x)

= 3.Log(2) / Log(2) → you can simplify by Log(2)

= 3

log(a) / log(b) is not a/b. If it was that easy, why would we use logarithms?

log(8) / log(2) =>
log(2^3) / log(2) =>
3 * log(2) / log(2)

The entire expression of log(t)/log(t) cancels out

3 * 1 =>
3

Remember 8 = 2 x 2 x 2 = 2^3
Hence
log8 = log2^3

log2^3 = 3log2 (The rules for indices and logarithms).
Hence
log 8 / log2 = 3log2/log2

Cancel down by 'log2'
Hence answer is '3' .

8 is 2^3 so by the rules of logs
log(8) = log[2^3} = 3log(2), and so,
log(8)/log(2) = 3

Extra note: A logarithm is a power.
For example when you write log(1000) = 3
that 3 is the power in 10^3 = 1000

The ratio of log(8) to log(2) is NOT the same as the ratio 8 to 2
As we have shown log(8)/log(2) = 3, whereas 8/2 = 4
Powers and simple division are different things