How do I solve $({log}_{3} 64) \cdot ({log}_{2} (\frac{1}{27}))$ $(log_3 64) *(log_2 (1/27))$ ?

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How do I solve $({log}_{3} 64) \cdot ({log}_{2} (\frac{1}{27}))$ $(log_3 64) *(log_2 (1/27))$ ?

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Answer 1 · 2017-10-14T18:10:25

$- 18$ $-18$

Explanation:

We should see that the bases are mismatched:

$64 = 2^{2 \cdot 3}$ $64=2^(2*3)$

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

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

We need to change ${log}_{3} x$ $log_3x$ to ${log}_{2} x$ $log_2x$ and vice versa. So:
${log}_{3} (64) = \frac{{log}_{2} (64)}{{log}_{2} (3)}$ $log_3(64)=(log_2(64))/(log_2(3))$

${log}_{2} (\frac{1}{27}) = \frac{{log}_{3} (\frac{1}{27})}{{log}_{3} (2)}$ $log_2(1/27)=(log_3(1/27))/(log_3(2))$

${log}_{3} (64) \cdot {log}_{2} (\frac{1}{27}) = \frac{{log}_{2} (64)}{{log}_{2} (3)} \cdot \frac{{log}_{3} (\frac{1}{27})}{{log}_{3} (2)}$ $log_3(64)*log_2(1/27)=(log_2(64))/(log_2(3))*(log_3(1/27))/(log_3(2))$

Since ${log}_{a} (b) \cdot {log}_{b} (a) = 1$ $log_a(b)*log_b(a)=1$

$= \frac{6 \cdot - 3}{1}$ $=(6*-3)/1$

$= - 18$ $=-18$

Proving some things:

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

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

${log}_{b} (a^{y}) = {log}_{b} (x)$ $log_b(a^y)=log_b(x)$

$y {log}_{b} (a) = {log}_{b} (x)$ $ylog_b(a)=log_b(x)$

$y = \frac{{log}_{b} (x)}{{log}_{b} (a)}$ $y=(log_b(x))/(log_b(a))$

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

${log}_{a} (x) = \frac{{log}_{b} (x)}{{log}_{b} (a)}$ $log_a(x)=(log_b(x))/(log_b(a))$

$Q . E . D .$ $Q.E.D.$

Let's prove that ${log}_{a} (b) \cdot {log}_{b} (a) = 1$ $log_a(b)*log_b(a)=1$
Let $a^{x} = b, b^{y} = a$ $a^x=b, b^y=a$ prove $x y = 1$ $xy=1$ so $y = \frac{1}{x}$ $y=1/x$
Therefore with $a^{x} = b$ $a^x=b$ prove $b^{\frac{1}{x}} = a$ $b^(1/x)=a$

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

change base from $b$ $b$ to $a$ $a$ :

${log}_{b} (a) = \frac{{log}_{a} (a)}{{log}_{a} (b)} = \frac{1}{{log}_{a} (b)}$ $log_b(a)=(log_a(a))/(log_a(b))=1/(log_a(b))$

Since we know that ${log}_{a} (b) = x$ $log_a(b)=x$
${log}_{b} (a) = \frac{1}{x}$ $log_b(a)=1/x$

$Q . E . D .$ $Q.E.D.$

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