ln(x^2y)^3. when you expand this natural log is e. so it would be 6lnex +3lney. but my answer doesn't have an e in it. how come?

when you write ln instead of log, the base e is implied.

if i wrote e, is it technically wrong?

Not really wrong.

To understand why your expanded expression does not include an "e", let's go through the steps of expanding the natural logarithm expression ln((x^2y)^3).

Step 1: Apply the power rule of logarithms.
ln((x^2y)^3) = 3ln(x^2y)

Step 2: Apply the product rule of logarithms.
3ln(x^2y) = 3[ln(x) + ln(y^2)]

Step 3: Apply the power rule of logarithms again.
3[ln(x) + ln(y^2)] = 3ln(x) + 3ln(y^2)

Step 4: Simplify the expression.
3ln(x) + 3ln(y^2) = ln(x^3) + ln(y^6)

So, the correct expanded form of ln((x^2y)^3) is ln(x^3) + ln(y^6). As you can see, there is no "e" in this expression.

It seems like there may be some confusion regarding the base of the natural logarithm. The natural logarithm, denoted as ln(x), has a base of "e" (Euler's number), which is an irrational constant approximately equal to 2.71828.

If you want to include "e" in your expanded expression, you would need to evaluate ln(x^3) and ln(y^6) using the properties of logarithms and the value of "e", resulting in a numerical approximation. But in the general form, ln(x) does not inherently involve "e", and its expansion may not necessarily contain "e" explicitly.