Question

What is 2log5(5x^3)+(1/3)log5((x^2)+6) written as a single logarithm?

Answer 1

Assuming 5 is the base. I'm going to leave that out for now.

2log(5x^3) + (1/3)log(x^2+6)

power rule
log(5^2 x^3*2) + log((x^2 + 6)^(1/3))

log(25x^6) + log((x^2 + 6)^(1/3))

quotient rule
log(25x^6 / (x^2 + 6)^(1/3))

Answer 2

Answer: =$log_5\ (25\ x^6\ \sqrt[3]{x^2+6})$

Step-by-step explanation:

Given log expression  $2log_5(5x^3)+\frac{1}{3} log_5((x^2)+6)$.

First we would apply log rule of exponents : $nlog_b(a) = log_b(a)^n$

$2log_5(5x^3)+\frac{1}{3} log_5((x^2)+6) = log_5(5x^3)^2+ log_5((x^2)+6)^{\frac{1}{3}}$.

Converting rational exponent into radical form, we get

$log_5(5x^3)^2+ log_5\sqrt[3]{((x^2)+6)}$.

Simplifying $(5x^3)^2 = 5^2x^{3\times 2} = 25x^6$

= $log_5(25x^6)+ log_5\sqrt[3]{((x^2)+6)}$.

Applying sum of logs rule $log_b(m)+log_b(n) = log_b(m\times n)$.

=$log_5\ (25\ x^6\ \sqrt[3]{x^2+6})$