Solve $2\ln(3x+5)=4$
The second method is:
$\ln(3x+5)^2=4$,
$\implies \ln(3x+5)=2$,
$\implies e^2=3x+5$,
$\implies x= \frac{e^2-5}{3}$
which equals only $ 0.796$, and there is not another answer (assuming if it is not rejected) This is not an exam's question; I already have the answer and know it, but what I am asking now is that: when I squared both sides, am I considered to be canceling another solution, which is the $-4.13$ (assuming if it was not to be rejected as it is known that no $\ln$ for negative)? And if yes, I am considered to be canceling a solution; why is this possible? And here is a picture of the whole thing I explained to be clearer.
I don't know any situation in which I would say that squaring an equation "cancels" a solution. It sometimes introduces extraneous "solutions" that you have to reject.
I suppose you could say that squaring sometimes forces you to "cancel a solution" later in the process.
In this case the rule "there is no $\ln$ of a negative number" does not apply, since the thing you apply $\ln$ to is $3x + 5,$ which is positive as long as $x > -5/3.$ Instead, you reject the negative square root because it implies that the output of $\ln$ is negative, and in this case you know that $\ln(3x+5)>0$ since it is multiplied by a positive number and produces a positive number.
So in the end you do "cancel a solution", not at the time when you square $\ln(3x+5),$ but rather at the time when you take the square root of $4$ and write only the positive square root.