Question

Question #5208b

Answer 1

I would say FALSE.

Explanation:

Consider it:
`e^(6lnx)=`
let us focus our attention on the exponent. We can use the property of logs to write it as:
`=e^(lnx^6)=`
now we use the definition of log and the fact that `e` and `ln` eliminate each other to give: `x^6`, or:
`=cancel(e)^(cancel(ln)x^6)=x^6`

Answer 2

False.

Explanation:

`e^(6lnx)=x^6`, not `6x`, for the following reason.

Recall the following property of logs:
`alnx=lnx^a`

That means `6lnx` is equivalent to:
`lnx^6`

But, since `e^x` and `lnx` are inverses, `e^lnx=x`. Likewise, `e^(lnx^6)=x^6`.

Note
Because `e^(6lnx)` isn't defined for `x<=0` (meaning if you plugged in a negative number for `x` you would get "ERROR" on your calculator), its equivalent of `x^6` is also not defined for `x<=0`. That means we have to limit the `x` values to `0` or positive numbers, so we write:
`e^(6lnx)=x^6` for `x>=0`

Answer 3

`x^6 ne 6x`

Explanation:

`e^(6lnx)=e^{log_e x^6}= x^6 ne 6x`