Is $f(x) = 2 + \ln x$ another way to write $f(x) =\log_e x +2$?

Question

I just want to make sure I am correctly understaning this concept.

$f(x) = 2 + \ln x$ is the same as $f(x) =\log_e x +2$

Thus my T graph would look like so:

e^y|x+2

-3|2.049

-2|2.135

-1|2.36

Is this correct?

Answer 1

Facts \[ y=2+\ln x \] \[ e^{y}=e^{2+\ln x}=e^{2}e^{\ln x}=e^{2}x \] T-Graph \[ \begin{array}{c|l} x & e^{2}x \\ \hline e^{-1} & e^{2}e^{-1} =e^{1} \\ e^{-2} & e^{2}e^{-2}=1 \\ e^{-3} & e^{2}e^{-3}=e^{-1} \\ \end{array} \]

Answer 2

Okay. Contrary to what people have stated it is NOT correct to say $f(x)=2+ln(x)= log_e(x+2)$. The miscommunication on the communities part is that it is correct that $f(x)=2+ln(x) = 2+ log_e(x)$. It is very important to indicate what the logarithm is acting on! That is why we use parentheses. Without them, we indicate that the logarithm affects only the first letter, variable, number, etc. and the rest is continuing with order of operations.

So the T-chart that you made is not correct, but it is a good step. Instead, replace it with $e^y$ and $e^{(log_e(x)+2)}=e^{log_e(x)}*e^2 = x*e^2$. Try to replace these values in your column.