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)+(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)+(1)/(3) log_5((x^2)+6) = log_5(5x^3)^2+ log_5((x^2)+6)^{(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* 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* n)`.

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