Answer:
Considering the Product and the Power rule of the Logarithms, they're both equivalent for x >0
Step-by-step explanation:
1) Considering the Product and the Power rule of the Logarithms:
[tex]\\log_{c}(ab)=log_{c}a+log_{c}b[/tex]
[tex]\\log_{c}a^{b}=blog_{c}[/tex]
2) Therefore we can say that:
[tex]\\f(x) = ln ((x)(x^2 + 1))^{\frac{1}{2}}\\\Rightarrow f(x)=ln (x(x^2 + 1))^{\frac{1}{2}}\Rightarrow f(x)=\frac{1}{2}ln((x)(x^{2} + 1)).\\g(x) = \frac{1}{2}(ln (x) + ln(x^{2} + 1))\Rightarrow g(x)=\frac{1}{2}ln(x(x^2+1))[/tex]
