Alg9.4.10

Find the value of $\log_3(\sqrt{243})\,$

Expressing the loga number in terms of the base,we get

$\log_3(\sqrt{243})=\log_3(\sqrt{81\cdot3})\,$

Simplifying

$\log_3(9\sqrt{3})\,$

$\log_3(3^{2+\frac{1}{2}})\,$

$\log_3(3^{\frac{5}{2}})\,$

This can be expressed as

$\frac{5}{2}\log_3(3)\,$

$\frac{5}{2}\cdot1\,$

Hence

$\log_3(\sqrt{243})=\frac{5}{2}\,$