How do you calculate ${log}_{5} (4)$ $log_5(4)$ ?

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Answer 1 · 2016-04-08T07:08:49

${log}_{5} (4) = 0.8614$ $log_5(4)=0.8614$

Let ${log}_{b} a = x$ $log_ba=x$ , then $b^{x} = a$ $b^x=a$ .

If $a = 10^{n}$ $a=10^n$ and $b = 10^{m}$ $b=10^m$ , then $n = log a$ $n=loga$ and $m = log b$ $m=logb$ and

$b^{x} = a$ $b^x=a$ becomes ${(10^{m})}^{x} = 10^{n}$ $(10^m)^x=10^n$ or $10^{m x} = 10^{n}$ $10^(mx)=10^n$ i.e.

$m x = n$ $mx=n$

Hence $x = \frac{n}{m} = \frac{log a}{log b}$ $x=n/m=loga/logb$

Thus ${log}_{5} (4) = \frac{log 4}{log 5} = \frac{0.6021}{0.6990} = 0.8614$ $log_5(4)=log4/log5=0.6021/0.6990=0.8614$

How do you calculate log_5(4) log5(4) log_5(4) ?