Cansomeone explain why log base a of B + log base b of C = log base a of C. I've asked all tutors and no help.

Answer 1

log[a] B + log[b] C

Using the change of bases rule:
log[b] C = lob[a]C / log[a]B

So we have
log[a]B + log[a]C / log[a]B

Now let's get the same denominator:
(log[a]B)² / log[a]B + log[a]C / log[a]B

Add the fractions:
(log[a]B)² + log[a]C) / log[a]B

Hey what's going on here? This isn't going to reduce correctly?!?

Let's try some numbers and confirm this. For example, let A = B = C = 10

log(10) + log(10) = log(10) ???

Nope there is something wrong with how someone tried to apply the rules of logs or wrote the question.

I think you must have meant *times*:
log[a]B *times* log[b]C = log[a]C

This does come directly from the change of bases rule.
Using the logic from above:
log[a]B * log[a]C / log[a]B --> cancel out the log[a]B and you have:

log[a]C

So if you correct your equation to be *times* rather than plus:
log[a]B * log[b]C = log[a]C

As a double check, let's try some numbers:
A = 2, B = 4, C = 64

log[2]4 = 2
log[4]64 = 3
log[2]64 = 6
2 x 3 = 6

Answer 2

That is true, I'm a tutor and it's hard to describe why it works. Take log base 2 8 (which equals 3) and log base 8 64 (which equals 2). This does equal log base 2 64 (5)

Answer 3

Well, it's a property of logarithms.

If you need the proof, here it is:

Suppose r = logbaseb u and s = logbaseb v,

u = b^r and v = b^s

u*v = b^r * b^s
uv = b^(r+s) ---- properties of exponents

the logarithm of this equation is equal to:

Since r + s = logbaseb u + logbaseb v then

logbaseb uv = logbaseb u + logbaseb v.