Why is

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

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

so on and so forth?

Results 1 to 6 of 6

- Sep 7th 2010, 05:43 AM #1

- Joined
- Aug 2010
- From
- Singapore
- Posts
- 93

- Sep 7th 2010, 05:50 AM #2

- Sep 7th 2010, 06:01 AM #3

- Joined
- Aug 2010
- From
- Singapore
- Posts
- 93

- Sep 7th 2010, 06:04 AM #4

- Joined
- Aug 2010
- From
- Singapore
- Posts
- 93

- Sep 7th 2010, 06:08 AM #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 $\displaystyle \log_e$.

- Sep 7th 2010, 12:37 PM #6
Duplicate thread: http://www.mathhelpforum.com/math-he...tc-155461.html

Thread closed.