Solve $\text {ln x} = \text {ln 1}$ (read before you answer)

Question

BEFORE YOU ANSWER:

$\text {ln x} = \text {ln 1}$

If I ask you to solve for x, even myself, would simply say:

ln x = ln 1 $\to$ x = 1

But,

If I were to solve the right hand side first (since it does not contain a variable), I would get:

ln x = $0$

This would mean x = -$\infty$.

Depending on the way you solve it, it gives you two different answers. Which is the right answer? I've been doing it the first way for so long, because I don't know why I thought of the second way today, and it's really confusing to what the right answer is?

Answer 1

How did you conclude $\ln x=0\implies x=-\infty$? No, it is not true. $e^x=0$ implies $x=-\infty$, not logarythm.

Answer 2

Exponentiate both sides to get $x=1$ since $e^{\ln x}=x$ and $e^0=1$

Answer 3

A logaritm says: to what power (x) do I have to raise a number to get the value that I know.

For example: ln(x) = 5, means: e^5 = x

So, ln(1) = 0, because if you raise e to the power of 0 you get 1.

ln(x) = ln(1) = 0 -> x = 1, because if you raise e to the power of 0, you get 1.

When you have a form ln(x) = y (where y is a known number), it's easy to solve because you just raise e to the power of y to get x

Sorry for my bad english, my math-english is not very accurate.

Hope this helped :)

Answer 4

Perhaps it is helpful to look at the graph of lnx.

If we look at this graphically, we see that lnx=0 when x=1. I think you are considering the wrong axis.

Answer 5

As a real-valued function of a real variable, the logarithmic function $\ln x$ is only defined for positive values of $x$. It's true that $\lim_{x\to0^+}\ln x=-\infty$, which is sometimes abbreviated as $\ln(0)=-\infty$. It looks like you are thinking of that abbreviation but getting it backwards, as $\ln(-\infty)=0$.