SOLUTION: Hi, please could you give me a hand on these questions? I have tried hard but i can't find my way with proofs: 3i)Prove that log of m (in the base a)

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Question 72121: Hi, please could you give me a hand on these questions? I have tried hard but i can't find my way with proofs:
3i)Prove that log of m (in the base a) - log of n (in the base a) = log of (m over n)( in the base a).
4i)Prove that nlog of x (in the base m) = log of x to the power of n (in the base m).
5a)Prove that log of b (in the base a) = log of b( in the base 10) over log of a (in the base 10).
Thank you very much for your help!!!!
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
let log(base a)m = X ; let log(base a)n= Y
Then a^(X)=m and a^Y=n
Then m/n = a^X/a^Y = a^(X-Y)
Take the log(base a) of both sides to get:
log(base a)(m/n) = log(base a)a^(X-Y)
log(base a)(m/n) = X-y
log(base a)(x/n) = log(base a)m - log(base a)n
==============
The proof for log(base a)mn = log(base a)m + log(base a)n
is almost identical:
You get a line that says:
mn=a^X * a^Y = a^(X+Y)
Take the log(base a) of both sides to get:
log(base a)mn = log(base a)a^(X+Y)
log(base a)mn = X+Y
log(base a)mn = log(base a)m + log(base a)n
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log(base a)m^n = ?
Let log(base a)m = X
Then a^X = m
Substitute for "m" to get:
log(base a)(a^X)^n = log(base a)a^(Xn)= Xn
Substitute for X to get:
=nlog(base a)m
----------------
Log(base a)b = x
Let a^x = b
Then x log(base 10) a = log(base 10 b
And x= [log(base 10)b] / [log(base 10)a]
Substitute for "x" to get
log(base a)b = log(base 10)b /log(base 10)a
-----------
Cheers,
Stan H.

SOLUTION: Hi, please could you give me a hand on these questions? I have tried hard but i can't find my way with proofs: 3i)Prove that log of m (in the base a) - log of n (in the base a)