Question

Hello, I need help with question #7Only part 2)A) and B)

Answer 1

The best way to find Arianna's errors is to attempt to work out the problem on our own.

We are given the expression

`\ln{(√(a))/(e^4b)}`

Her first step was correct, we should rewrite our radical as a fractional exponent.

`\ln{\frac{a^{(1)/(2)}}{e^4b}}`

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:

`\ln{(x)/(y)}=ln(x)-ln(y)`

Instead, we will have

`\ln{a^{(1)/(2)}}-ln(e^4b)`

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:

`ln(xy)=ln(x)+ln(y)`

So, we will have

`\ln{a^(1)/(2)}-(ln(e^4)+ln(b))`

Applying the product property, Arianna should have moved the 4 to be a coefficient of ln(e):

`(1)/(2)ln(a)-(4ln(e)+ln(b))`

The rest of her work is fine, so let's just finish the expression from here. The identity property of natural states

`ln(e)=1`

so now we can have

`\begin{gathered} (1)/(2)ln(a)-(4\cdot1+ln(b)) \\ (1)/(2)ln(a)-(4+ln(b)) \end{gathered}`

And finally, we will distribute the negative:

`(1)/(2)ln(a)-4-ln(b)`

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

`(1)/(2)ln(a)-4-ln(b)`