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

May 29, 2015

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

${\log}_{a} \left(x\right)$ (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,

${a}^{{\log}_{a} \left(x\right)} = x$

As is well known from the properties of the exponential function, ${a}^{p + q} = {a}^{p} \cdot {a}^{q}$
Using this for $p = {\log}_{a} \left(x\right)$ and $q = {\log}_{a} \left(y\right)$, we get the following:

${a}^{{\log}_{a} \left(x\right) + {\log}_{a} \left(y\right)} = {a}^{{\log}_{a} \left(x\right)} \cdot {a}^{{\log}_{a} \left(y\right)} = x \cdot y = {a}^{{\log}_{a} \left(x \cdot y\right)}$

Therefore,
${\log}_{a} \left(x\right) + {\log}_{a} \left(y\right) = {\log}_{a} \left(x \cdot y\right)$

As a consequence from this,
${\log}_{a} \left(x\right) + {\log}_{a} \left(\frac{1}{x}\right) = {\log}_{a} \left[x \cdot \left(\frac{1}{x}\right)\right] = {\log}_{a} \left(1\right) = 0$
Therefore,
${\log}_{a} \left(\frac{1}{x}\right) = - {\log}_{a} \left(x\right)$
From this we also can derive the following:
${\log}_{a} \left(\frac{x}{y}\right) = {\log}_{a} \left[x \cdot \left(\frac{1}{y}\right)\right] = {\log}_{a} \left(x\right) + {\log}_{a} \left(\frac{1}{y}\right) =$

$= {\log}_{a} \left(x\right) - {\log}_{a} \left(y\right)$

Using all this theory, we can calculate
${\log}_{8} \left(7\right) - {\log}_{8} \left(s\right) + {\log}_{8} \left(t\right) - {\log}_{8} \left(4\right) =$
$= {\log}_{8} \left(\frac{7}{s}\right) + {\log}_{8} \left(\frac{t}{4}\right) = {\log}_{8} \left(\frac{7 t}{4 s}\right)$