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.

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 --%26gt; 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

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.
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)
Reply: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.

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