# Thread: Finding X in the Exponent

1. ## Finding X in the Exponent

I am stuck on these three problems. I have to find out what x is. Any help would be greatly appreciated.

$1.06^x=4.1$

$50e^{.035x}=200$

$3+2e^{-x}=6$

Thank you!

2. Hmm...I'm not sure if this is right but I'd probably take the natural log of both sides.

so i should divide $\frac{ln{4.1}}{ln{1.06^x}}$ but that gives me 24.21609205, which could not be X. or did i do this wrong? same thing for the other problems?

4. Originally Posted by OnMyWayToBeAMathProffesor
so i should divide $\frac{ln{4.1}}{ln{1.06^x}}$ but that gives me 24.21609205, which could not be X. or did i do this wrong? same thing for the other problems?
No it should turn out to be

$ln1.06^x = ln4.1$

$(x)ln1.06 = ln4.1$ because of the power rule

$x = \frac {ln4.1}{ln1.06}$

and just use a calculator to solve that. Well it did come out to be the same answer. Idk because I'm not a helper, so you should ask someone else.

i am sorry, i did not mean to type the X in the ln equation. i did $\frac{ln{4.1}}{ln{1.06}}$ and got 24.21509205. Thanks for your help anyway.

Anybody else know?

6. Originally Posted by OnMyWayToBeAMathProffesor
i am sorry, i did not mean to type the X in the ln equation. i did $\frac{ln{4.1}}{ln{1.06}}$ and got 24.21509205. Thanks for your help anyway.

Anybody else know?
I checked and it seems to be the correct answer.

so if X=24.21609205 than $1.06^{24.21609205}$ should equal 4.1 but actually equals 1.079758817.

any ideas?

8. Originally Posted by OnMyWayToBeAMathProffesor
so if X=24.21609205 than $1.06^{24.21609205}$ should equal 4.1 but actually equals 1.079758817.

any ideas?
Really? I get approximately 4.1

o, i forgot to add the parenthesis. Thank you very much for all of your gracious help. Would i employ the same technique to find out the other 2,

$50e^{.035x}=200$

$3+2e^{-x}=6$

Once again, many thanks.

10. Originally Posted by OnMyWayToBeAMathProffesor
o, i forgot to add the parenthesis. Thank you very much for all of your gracious help. Would i employ the same technique to find out the other 2,

$50e^{.035x}=200$

$3+2e^{-x}=6$

Once again, many thanks.
Yes it should work too. Just remember that $ln(e) = 1$

11. thank you.