# Thread: [SOLVED] Natural Base e

1. ## [SOLVED] Natural Base e

I know that $e$ can be approximated by the form $1 + \frac {1}{1!} + \frac {2}{2!} + \frac {3}{3!} + \frac {4}{4!} +... \frac {n}{n!}$ .

Would $\frac {2^1}{1!} + \frac {2^2}{2!} + \frac {2^3}{3!} + \frac {2^4}{4!} +... \frac {2^n}{n!} = e^2$ ?

And how would you determine an aproximate value for $\frac {a^1}{1!} + \frac {a^2}{2!} + \frac {a^3}{3!} + \frac {a^4}{4!} +... \frac {a^n}{n!}$ ?

2. Yep:

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{n}}{n!} + ...$

If x = 2, you get what you get:
$e^{2} = 1+ 2 + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...$

If you have x = a, you get ... ?

3. Originally Posted by o_O
Yep:

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{n}}{n!} + ...$

If x = 2, you get what you get:
$e^{2} = 1+ 2 + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...$

If you have x = a, you get ... ?
$e^a$ right?

Edit - Two more thing. I know that this might be off topic...but since you answered my other thread too...and I don't want to make another thread to ask 2 simple questions....

What values of $x$ would make $\log x>\ln x$ and how?

$\log_3 8 + \log_3 y = 4\log_3 (2x)$ . If I wanted to solve this for "y" and I did, would this be right... $y=2x^4$ after condensing it and getting rid of log?

4. Hello,

Originally Posted by chrozer
$e^a$ right?
Yep

What values of $x$ would make $\log x>\ln x$ and how?
If $\log=\ln_{10}$ then $\log x = \frac{\ln x}{\ln{10}}$
If inverted, then...just invert it ^^'
Assuming that $\ln{10}>1$ (or $\log{10}$), can you conclude ?

$\log_3 8 + \log_3 y = 4\log_3 (2x)$ . If I wanted to solve this for "y" and I did, would this be right... $y=2x^4$ after condensing it and getting rid of log?
I find the same ^^'

5. Originally Posted by Moo
Hello,

Yep

If $\log=\ln_{10}$ then $\log x = \frac{\ln x}{\ln{10}}$
If inverted, then...just invert it ^^'
Assuming that $\ln{10}>1$ (or $\log{10}$), can you conclude ?

I find the same ^^'
Sorry but I have no idea how to conclude that.

6. ## You have

Originally Posted by chrozer
Sorry but I have no idea how to conclude that.
$log_{10}(x)>ln(x)\Rightarrow{\frac{ln(x)}{ln(10)}> ln(x)}=0\Rightarrow{ln(x)\bigg(\cdot{\frac{1}{ln(1 0)}-1}\bigg)>0}$ then using logic we see that $log_{10}(x)>ln(x),\forall{x}\in(0,1)$

7. Originally Posted by Mathstud28
$log_{10}(x)>ln(x)\Rightarrow{\frac{ln(x)}{ln(10)}> ln(x)}=0\Rightarrow{ln(x)\bigg(\cdot{\frac{1}{ln(1 0)}-1}\bigg)>0}$ then using logic we see that $log_{10}(x)>ln(x),\forall{x}\in(0,1)$
Ok....thnx alot.

8. Originally Posted by chrozer
$\log_3 8 + \log_3 y = 4\log_3 (2x)$ . If I wanted to solve this for "y" and I did, would this be right... $y=2x^4$ after condensing it and getting rid of log?
$\log_3 8 + \log_3 y = 4\log_3 (2x)$

$\log_3 (8y) = \log_3 (2x)^4$

$8y = 16x^4$

$y = 2x^4$