How do you calculate ${log}_{3} \sqrt{243}$ $log_3 sqrt243$ ?

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Answer 1 · 2016-04-24T01:38:25

First, use the exponent rule $\sqrt[a]{n} = n^{\frac{1}{a}}$ $root(a)(n) = n^(1/a)$

${log}_{3} (243^{\frac{1}{2}})$ $log_3(243^(1/2))$

Now, use the log rule ${log a}^{n} = n log a$ $loga^n = nloga$

$\frac{1}{2} {log}_{3} (243)$ $1/2log_3(243)$

Now, use the log rule ${log}_{a} n = \frac{log n}{log a}$ $log_an = logn/loga$

$\frac{1}{2} (\frac{log 243}{log 3})$ $1/2(log243/log3)$

Rewrite 243 in base 3.

$\frac{1}{2} (\frac{{log 3}^{5}}{log 3})$ $1/2((log3^5)/log3)$

$\frac{1}{2} (\frac{5 log 3}{log 3})$ $1/2((5log3)/log3)$

$\frac{1}{2} \times 5$ $1/2 xx 5$

$\frac{5}{2}$ $5/2$

Thus, completely simplified, ${log}_{3} (\sqrt{243}) = \frac{5}{2} or 2.5$ $log_3(sqrt(243)) = 5/2 or 2.5$

Hopefully this helps!

How do you calculate log3√243 log_3 sqrt243?