Logarithms!?

Question

Can ne 1 explain how u answer this question:
Given that x>0, y>0 show that
loga(x/y)=loga x-loga y
thanku for ne help available!
xxx

Answer 1

That's a standard relation, you don't have to prove that.

If you really want an explanation.
if f = loga(x/y)
g = loga(x)
h = loga(y)

Then x = a^g
y = a^h
x/y = a^g / a^h
= a^(g-h)
But x/y = a^f
So f = g - h
or loga(x/y) = loga(x) - loga(y)

Answer 2

Because (a^x)/(a^y) = a^(x - y)

Since the definition of loga(x) = y is a^y = x.

Answer 3

This is a basic property of logarithms.

Logarithms are exponents.. and as such.. they follow the same laws as exponents.

Exponent Law: When dividing with like bases, you subtract exponents..

This is reversed..

Logarithm Law: When subtracting with like bases, you divide.

Answer 4

log base a N= L, if a^L = N
Let M= a^x and N= a^y
Then by defnition of logarithms, we have:
x=log_a M and y = log_a N
Now M/N = a^x / a^y = a^(x-y)
Thus by definition of logarithms, log_a M/N= x-y, so we get the final result log_a M/N = log_aM-log_aN