Question 670957: Help with a proof for this: (log base a of x)/(log base a/b of x) = 1+(log base a of 1/b)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! log.a means log to the base of a (my interpretation not generally used).
your equation becomes:
log.a(x) / log.a/b(x) = 1 + log.a(1/b)
you want to prove this identity is true.
you need to use the log base conversion formula of log.d(x) = log.e(x) / log.e(d).
you use this conversion formula on log.a/b(x) to get:
log.a/b(x) = log.a(x) / log.a(a/b)
your equation of:
log.a(x) / log.a/b(x) = 1 + log.a(1/b) becomes:
log.a(x) / (log.a(x) / log.a(a/b)) = 1 + log.a(1/b)
since h/(i/j) is equivalent to h*(j/i), your equation becomes:
log.a(x) * log.a(a/b) / log.a(x) = 1 + log.a(1/b)
since log.a(x) / log.a(x) cancels out, your equation becomes:
log.a(a/b) = 1 + log.a(1/b)
since log(m/n) = log(m) - log(n), your equation becomes:
log.a(a) - log.a(b) = 1 + log.a(1) - log.a(b)
if you add log.a(b) to both sides of the equation, it will cancel out and you'll get:
log.a(a) = 1 + log.a(1)
since log.a(1) = 0, your equation becomes:
log.a(a) = 1
since log.a(a) = 1, your equation becomes:
1 = 1 which is true confirming the identity as being valid.
log.a(1) is equal to 0 based on the following logic.
log.a(1) = y if and only if a^y = 1
a^y = 1 if and only if y = 0
this makes log.a(1) = 0
log.a(a) is equal to 1 based on the following logic.
log.a(a) = y if and only if a^y = a
a^y = a if and only if y = 1
this makes log.a(a) = 1
the key to solving this is the use of the log base conversion formula which is.
log.k(x) = log.p(x) / log.p(k)
as an example, take log.2(8) = 3
this is true because 2^3 = 8
convert this to log.10
log.2(8) = log.10(8) / log.10(2)
use your calculator to see that the answer is going to be 3.
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