How do I solve (log_3 64) *(log_2 (1/27)) ?

Oct 14, 2017

$- 18$

Explanation:

We should see that the bases are mismatched:

$64 = {2}^{2 \cdot 3}$

$\frac{1}{27} = {3}^{-} 3$

So, we should apply this property of logs to change the bases:
${\log}_{a} x = \frac{{\log}_{b} \left(x\right)}{{\log}_{b} \left(a\right)}$
Where $a$ is the original base and $b$ is the new base.

We need to change ${\log}_{3} x$ to ${\log}_{2} x$ and vice versa. So:
${\log}_{3} \left(64\right) = \frac{{\log}_{2} \left(64\right)}{{\log}_{2} \left(3\right)}$

${\log}_{2} \left(\frac{1}{27}\right) = \frac{{\log}_{3} \left(\frac{1}{27}\right)}{{\log}_{3} \left(2\right)}$

${\log}_{3} \left(64\right) \cdot {\log}_{2} \left(\frac{1}{27}\right) = \frac{{\log}_{2} \left(64\right)}{{\log}_{2} \left(3\right)} \cdot \frac{{\log}_{3} \left(\frac{1}{27}\right)}{{\log}_{3} \left(2\right)}$

Since ${\log}_{a} \left(b\right) \cdot {\log}_{b} \left(a\right) = 1$

$= \frac{6 \cdot - 3}{1}$

$= - 18$

Proving some things:

Let's prove that ${\log}_{a} x = \frac{{\log}_{b} \left(x\right)}{{\log}_{b} \left(a\right)}$
Let's start with ${a}^{y} = x$

Take ${\log}_{b}$ of both sides, where $b$ is the target base:

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

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

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

We know that in ${a}^{y} = x$, $y = {\log}_{a} \left(x\right)$ therefore:

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

$Q . E . D .$

Let's prove that ${\log}_{a} \left(b\right) \cdot {\log}_{b} \left(a\right) = 1$
Let ${a}^{x} = b , {b}^{y} = a$ prove $x y = 1$ so $y = \frac{1}{x}$
Therefore with ${a}^{x} = b$ prove ${b}^{\frac{1}{x}} = a$

Rewrite with logs:
${\log}_{a} \left(b\right) = x$ and prove ${\log}_{b} \left(a\right) = \frac{1}{x}$

change base from $b$ to $a$:

${\log}_{b} \left(a\right) = \frac{{\log}_{a} \left(a\right)}{{\log}_{a} \left(b\right)} = \frac{1}{{\log}_{a} \left(b\right)}$

Since we know that ${\log}_{a} \left(b\right) = x$
${\log}_{b} \left(a\right) = \frac{1}{x}$

$Q . E . D .$