Respuesta :
Answer:
[tex]\ln x(x^2 + 1)^\frac{1}{3} = \ln x + \dfrac{1}{3} \ln (x^2 + 1)[/tex]
Step-by-step explanation:
We are given the following expression in the question
[tex]\ln x(x^2 + 1)^\frac{1}{3}[/tex]
Logarithmic Properties:
[tex]\log (ab) = \log a + \log b\\\\\log \dfrac{a}{b} = \log a - \log b\\\\\log (a^b) = b\log a[/tex]
We have to simplify the given expression
[tex]\ln x(x^2 + 1)^\frac{1}{3}\\=\ln x + \ln (x^2 + 1)^\frac{1}{3}\\\ln x + \dfrac{1}{3} \ln (x^2 + 1)[/tex]
[tex]\ln x(x^2 + 1)^\frac{1}{3} = \ln x + \dfrac{1}{3} \ln (x^2 + 1)[/tex]
Answer:
[tex]\ln x+\frac{1}{3}\ln (x^2 + 1)[/tex]
Step-by-step explanation:
Consider the given expression is
[tex]\ln x(x^2 + 1)^{\frac{1}{3}}[/tex]
We need to rewrite the expression as a sum,difference,or multiple of logarithms.
Using the properties of logarithm we get
[tex]\ln x+\ln (x^2 + 1)^{\frac{1}{3}}[/tex] [tex][\because \ln(ab)=\ln a+\ln b][/tex]
[tex]\ln x+\frac{1}{3}\ln (x^2 + 1)[/tex] [tex][\because \ln(a^b)=b\ln a][/tex]
Therefore, the equate form of given expression is [tex]\ln x+\frac{1}{3}\ln (x^2 + 1)[/tex] .
