How do you solve ${log}_{4} r = 3$ $log_4 r=3$ ?

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Answer 1 · 2016-10-08T08:16:15

$r = 64$ $r=64$

From the definition of logarithm, if

${log}_{4} r = 3$ $log_4 r=3$ , we have

$4^{3} = r$ $4^3=r$

Hence $r = 64$ $r=64$

Answer 2 · 2016-10-08T08:16:55

$r = 64$ $r=64$ .

Recall the Defn. of $log$ $log$ fun.

$Defn. : {log}_{b} x = m \Leftrightarrow b^{m} = x$ $" Defn. : "log_b x=m iff b^m=x$

$Hence, {log}_{4} r = 3 \Rightarrow 4^{3} = r$ $"Hence, "log_4r=3 rArr 4^3=r$

$:. r=64$ .

How do you solve log_4 r=3log4r=3log_4 r=3?