Answer:
[tex][\frac{ln x(x + 3)^2}{x^2 - 1}]^\frac{1}{3} [/tex]
Step-by-step explanation:
1/3[2 In(x + 3) + In x - In(x^2 - 1)]
[tex]\frac{1}{3} [2 ln(x + 3) + ln x - ln(x^2 - 1)][/tex]
m ln(x)= lnx^m
move the term before ln to the exponent
[tex]\frac{1}{3} [ln(x + 3)^2 + ln x - ln(x^2 - 1)][/tex]
ln m + ln n= ln(mn)
[tex]\frac{1}{3} [ln x(x + 3)^2- ln(x^2 - 1)][/tex]
ln m - ln n = ln(m/n)
[tex]\frac{1}{3} [\frac{ln x(x + 3)^2}{x^2 - 1}][/tex]
Now move fraction 1/3 to the exponent
[tex][\frac{ln x(x + 3)^2}{x^2 - 1}]^\frac{1}{3} [/tex]
