^{b}= x, then, log

_{a}x = b. John Napier was the first person to introduce logarithms in the 1600s and they rapidly came into use, they were especially useful for one fact about logarithms.

That is that log

_{a}(xy) = log

_{a}(x) + log

_{a}(y) . This made large multiplications a lot simpler and simply turned it into a problem of addition. And in fact we can prove this fact relatively simply, and I will do after proving one other property. That is that log

_{a}(x

^{c}) = c log

_{a}(x).

Let log

_{a}(x) = b, if we multiply both sides by c and write log

_{a}(x) = b in exponent form we get:

c log

_{a}(x) = bc and a

^{b}= x. If we raise both sides of a

^{b}= x to the power of c we get:

a

^{bc}= x

^{c }turning this into a log gives:

log

_{a}(x

^{c}) = bc, and we know that bc = c log

_{a}(x), so this means that:

log

_{a}(x

^{c}) = c log

_{a}(x), proving what we needed to know.

Now to prove that log

Let m = log

_{a}(xy) = log_{a}(x) + log_{a}(y):Let m = log

_{a}(x) and n = log_{a}(y) and then write these in exponent form
x = a

^{m }and y = a^{n }then we multiply these together to give:
x × y = a

^{m }× a^{n}= a^{m+n}, we can now take log_{a}of both sides and evaluate
log

_{a}(xy) = log_{a}(a)^{m+n}, then we use what we proved previously to get:
log

log

log

_{a}(xy) = (m + n) log_{a}a, note that log_{a}a = 1 for all a.log

_{a}(xy) = m + n, we know from our first line that m = log_{a}x and n = log_{a}y so:log

_{a}(xy) = log_{a}(x) + log_{a}(y), proving what we want!
We can utilise the last two properties to also prove another property (we can also use the last method to prove it too, but this is far easier).

Consider log

_{a}(x/y) and notice that this can be wrote as:
log

log

log

And that really is it for the basics of logarithms, you can do a lot with them so make sure you can understand these simply properties, they will make other problems a lot, lot simpler.

W4JUG97V4GUR

_{a}(x × y^{-1}), which from using our last fact means that:log

_{a}(x × y^{-1}) = log_{a}(x) + log_{a}(y^{-1}), using the first property (log_{a}(x^{c}) = c log_{a}(x)) this means that:log

_{a}(x/y) = log_{a}(x) - log_{a}(y), providing another identity.And that really is it for the basics of logarithms, you can do a lot with them so make sure you can understand these simply properties, they will make other problems a lot, lot simpler.

W4JUG97V4GUR

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