The best way to find Arianna's errors is to attempt to work out the problem on our own.
We are given the expression
[tex]\ln{\frac{\sqrt{a}}{e^4b}}[/tex]Her first step was correct, we should rewrite our radical as a fractional exponent.
[tex]\ln{\frac{a^{\frac{1}{2}}}{e^4b}}[/tex]However, in the next step, she should not have written the expression as two natural logs as a fraction. Instead, we have to apply the quotient property:
[tex]\ln{\frac{x}{y}}=\ln{x}-\ln{y}[/tex]Instead, we will have
[tex]\ln{a^{\frac{1}{2}}}-\ln{e^4b}[/tex]Arianna correctly rewrote the exponent of "a" as a coefficient, but not for "e". Before we get there, let's expand the second part of the expression using the product property:
[tex]\ln{xy}=\ln{x}+\ln{y}[/tex]So, we will have
[tex]\ln{a^\frac{1}{2}}-(\ln{e^4}+\ln{b})[/tex]Applying the product property, Arianna should have moved the 4 to be a coefficient of ln(e):
[tex]\frac{1}{2}\ln{a}-(4\ln{e}+\ln{b})[/tex]The rest of her work is fine, so let's just finish the expression from here. The identity property of natural states
[tex]\ln{e}=1[/tex]so now we can have
[tex]\begin{gathered} \frac{1}{2}\ln{a}-(4\cdot1+\ln{b}) \\ \frac{1}{2}\ln{a}-(4+\ln{b}) \end{gathered}[/tex]And finally, we will distribute the negative:
[tex]\frac{1}{2}\ln{a}-4-\ln{b}[/tex]Part 2 a) Arianna had these two mistakes: First, she should have written the division as a subtraction. Second, she should have written the exponent 4 as a coefficient in front of ln(e) rather than ln(4e).
Part 2 b) the final expression should really be
[tex]\frac{1}{2}\ln{a}-4-\ln{b}[/tex]