Why I get $e^0=-\inf$ !?

Question

I am trying to understand why my calculation for result of $$e^{0}=1$$ is not the same of others like posted here;

My calculation is based of using the $\ln$ and power actions which is shown below: $$$$

$$\text{If}\, e^{0}=y\overset{\ln}{\rightarrow} \ln(e^{0})=\ln(y)$$

$${\rightarrow} 0\times \ln(e)=\ln(y)\overset{}{\rightarrow} 0\times 1=\ln(y) \rightarrow 0=\ln(y)\rightarrow y =- \infty !$$

I guess my calculation s not correct by If seems correct, i guess it shows some strange aspect of e and power (^) by the blow

Thanks for your consideration.

Answer 1

You obviously confused logarithm and exponential functions. $\log(y)=0$ means $y=1$.

Answer 2

From your calculations

$\ln y=0 \Longrightarrow y=e^0=1$. This is a circular trivial manipulations. Because, you get $y=e^0$ again. I could not see why $y=- \infty?$.

Here, $y=e^0$ by using only the definition of logarithm. Just definition. Also, $e^0=1$ is based on exponential definition.

Answer 3

You are making a fundamental error. The logarithm of 0 is negative infinity, the logarithm of 1 is 0. From ln(y)= 0 you should conclude that y= 1, not negative infinity.