# Thread: Why is 10^lg 5=5 and e^ln5=5, etc?

1. ## Why is 10^lg 5=5 and e^ln5=5, etc?

Why is

$\displaystyle (10^\lg5)=5$
$\displaystyle (e^\ln5)=5$

so on and so forth?

2. Because the notation $\displaystyle \log$ is being used to represent the logarithm of base $\displaystyle 10$, and $\displaystyle \ln$ is being used to represent the logarithm of base $\displaystyle e$.

3. Originally Posted by Prove It
Because the notation $\displaystyle \log$ is being used to represent the logarithm of base $\displaystyle 10$, and $\displaystyle \ln$ is being used to represent the logarithm of base $\displaystyle e$.
Can you break it down further so that I can digest that information?

4. Thread closed. I dunno why i have double thread. Moderator can delete for me and transfer this reply over to the other?

5. You should know that exponentials undo logarithms and vice versa.

So $\displaystyle a^{\log_a{x}} = x$ and $\displaystyle \log_a{(a^x)} = x$.

In your first case, you're using $\displaystyle \lg$ to represent $\displaystyle \log_{10}$ and $\displaystyle \ln$ to represent $\displaystyle \log_e$.

6. Originally Posted by stupidguy
Thread closed. I dunno why i have double thread. Moderator can delete for me and transfer this reply over to the other?