How do you evaluate ${log}_{16} 1$ $log_[16]1$ ?

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How do you evaluate ${log}_{16} 1$ $log_[16]1$ ?

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Answer 1 · 2016-03-22T16:45:22

${log}_{16} 1 = 0$ $log_16 1=0$

Explanation:

${log}_{16} 1$ $log_16 1$
$= \frac{log 1}{log 16}$ $=log 1/log 16$ -> use change of base property ${log}_{b} x = \frac{log x}{log b}$ $log_bx=logx/logb$
$= \frac{0}{log 16}$ $=0/log 16$ -> $log 1 = 0$ $log 1=0$
$= 0$ $=0$

Answer 2 · 2016-03-22T17:22:27

0

Explanation:

By definition $log 1$ $log 1$ to any base is equal to $0$ $0$ .

Proof:
Let ${log}_{16} 1 = n$ $log_16 1=n$
By definition of logarithmic function

$16^{n} = 1$ $16^n=1$ , This is true only if $n = 0$ $n=0$

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