Question #5208b

Aug 16, 2016

I would say FALSE.

Explanation:

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

Aug 16, 2016

False.

Explanation:

${e}^{6 \ln x} = {x}^{6}$, not $6 x$, for the following reason.

Recall the following property of logs:
$a \ln x = \ln {x}^{a}$

That means $6 \ln x$ is equivalent to:
$\ln {x}^{6}$

But, since ${e}^{x}$ and $\ln x$ are inverses, ${e}^{\ln} x = x$. Likewise, ${e}^{\ln {x}^{6}} = {x}^{6}$.

Note
Because ${e}^{6 \ln x}$ isn't defined for $x \le 0$ (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 $x \le 0$. That means we have to limit the $x$ values to $0$ or positive numbers, so we write:
${e}^{6 \ln x} = {x}^{6}$ for $x \ge 0$

Aug 16, 2016

${x}^{6} \ne 6 x$

Explanation:

${e}^{6 \ln x} = {e}^{{\log}_{e} {x}^{6}} = {x}^{6} \ne 6 x$