How do you write ${log}_{4} (\frac{1}{16}) = - 2$ $log_4 (1/16) = -2$ into its exponential form?

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How do you write ${log}_{4} (\frac{1}{16}) = - 2$ $log_4 (1/16) = -2$ into its exponential form?

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Answer 1 · 2015-02-13T17:58:13

You use the definition of logarithm:
${log}_{a} x = b \Rightarrow a^{b} = x$ $log_ax=b => a^b=x$

So you can write your expression. taking the base $a$ $a$ of the log (in this case is $4$ $4$ ) to the power of $b = - 2$ $b=-2$ which is indeed equal to the argument of the log, i.e.:

$4^{- 2} = \frac{1}{4^{2}} = \frac{1}{16}$ $4^(-2)=1/4^2=1/16$

In which you use the fact that: $x^{- a} = \frac{1}{x^{a}}$ $x^-a=1/x^a$

Alternatively you can take all as exponents of $4$ $4$ :
$4^{{log}_{4} (\frac{1}{16})} = 4^{- 2}$ $4^(log_4(1/16))=4^(-2)$
Which is again:
$\frac{1}{16} = \frac{1}{16}$ $1/16=1/16$

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How do you write ${log}_{4} (\frac{1}{16}) = - 2$ $log_4 (1/16) = -2$ into its exponential form?

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