How do you evaluate ${log}_{81} 3$ $log_81 3$ ?

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How do you evaluate ${log}_{81} 3$ $log_81 3$ ?

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Answer 1 · 2016-11-02T22:08:36

${log}_{81} 3 = \frac{1}{4}$ $log_81 3 = 1/4$

Explanation:

${log}_{81} 3 = x$ $log_81 3 =x$

Rewrite as an exponential. Remember, the answer to a log is the exponent. In this case $x$ $x$ is the exponent, and $81$ $81$ is the base.

$81^{x} = 3$ $81^x =3$

Find a common base for both sides, which is $3$ $3$ .
$81 = 3^{4}$ $81=3^4$ , so by substitution...

${(3^{4})}^{x} = 3$ $(3^4)^x=3$

Use the exponent rule ${(x^{a})}^{b} = x^{a b}$ $(x^a)^b=x^(ab)$

$3^{4 x} = 3^{1}$ $3^(4x)=3^1$

$4 x = 1$ $4x=1$

$x = \frac{1}{4}$ $x=1/4$

Answer 2 · 2016-11-02T22:12:27

${log}_{81} (3) = \frac{1}{4}$ $log_81(3)=1/4$

Explanation:

Since $81$ $81$ is much larger than $3$ $3$ , our answer will be a decimal, so let's think of this problem in the opposite sense: ${log}_{3} (81)$ $log_3(81)$ . $3^{4} = 81$ $3^4 = 81$ , so ${log}_{3} (81) = 4$ $log_3(81) = 4$ .

Using the law of exponents, we know that if $a^{m} = n$ $a^m = n$ , then $n^{\frac{1}{m}} = a$ $n^(1/m) = a$ . So using this rule, we know that $81^{\frac{1}{4}} = 3$ $81^(1/4) = 3$ , so ${log}_{81} (3) = \frac{1}{4}$ $log_81(3)=1/4$

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How do you evaluate ${log}_{81} 3$ $log_81 3$ ?

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