why is the addition of same-base logarithms the same as the multiplication of same-base logarithms, and same for subtracting and dividing same-base logs?

Question

I know it’s because of the “product property” but that doesn’t explain the inner workings for me and the proof is bouncing right off my skull.

In: Mathematics

Answer 1

A logarithm is the inverse of n^x. So logn(n^x) = x. Kinda like how multiplication and division are inverses or addition and subtraction are inverses. So n^x and logn(x) follow the same rules.

Answer 2

You know how division undoes multiplication? That is 2*3 = 6 <-> 6/2 = 3. Or in general if x*y = z, then z/x = y.

Logarithms are one way of undoing exponents. So, 2^3 = 8 <-> log_2(8) = 3. And in general, if x^y = z then log_x(z) = y.

One useful fact about exponents is that a^b * a^c = a^(b+c). It’s easy to see why this works if you just expand out the terms. a^b = a*a*…*a b times. a^c = a*a* … *a c times. So when you multiply a^b and a^c it’s the same as a*a*…*a b + c times.

Let’s say that a^b = x and a^c = y. Then a^(b+c) = x*y.

And logarithms undo exponents, so log_a(x) = b and log_a(y) = c. and log_a(x*y) = b + c = log_a(x) + log_a(y).