Question

How do you simplify `log_8 7 – log_8 s + log_8 t – log_8 4`?

Answer 1

Let's revisit the theory of logarithms before we address this problem.

`log_a(x)` (where a>0 and x>0) is a number that, if used as an exponent with a base `a`, produces `x`.

That is, by definition of the logarithm,

`a^(log_a(x)) = x`

As is well known from the properties of the exponential function, `a^(p+q)=a^p*a^q`
Using this for `p=log_a(x)` and `q=log_a(y)`, we get the following:

`a^[log_a(x)+log_a(y)] = a^[log_a(x)] * a^[log_a(y)] = x*y = a^[log_a(x*y)]`

Therefore,
`log_a(x)+log_a(y) = log_a(x*y)`

As a consequence from this,
`log_a(x)+log_a(1/x) = log_a[x*(1/x)] = log_a(1) = 0`
Therefore,
`log_a(1/x) = -log_a(x)`
From this we also can derive the following:
`log_a(x/y) = log_a[x*(1/y)] = log_a(x)+log_a(1/y) =`

`= log_a(x)-log_a(y)`

Using all this theory, we can calculate
`log_8(7)-log_8(s)+log_8(t)-log_8(4) =`
`= log_8(7/s)+log_8(t/4) = log_8((7t)/(4s))`