quick math help please

[OogyWaWa]

SOLVED! Thanks all!

trying to figure out some finance homework and I don't know how to factor this out. how do you factor out variable exponents?

ex)

3200 * (1.12)^n = 100000
solve for n

do you have to do some kind of log to get the N out?

 

[RESmonkey]

Finance lol

Use natural logarithm --> ln()

 

[MotF Bane]

Get the 3200 on the same side as the 100000, then take logs.

 

[OogyWaWa]

Get the 3200 on the same side as the 100000, then take logs.

logs of?

 

[Hacp]

Divide and take the natural log of both sides. Are you sure this is finance homework and not high school algebra?

 

[jchu14]

and keep in mind of the log property:

ln(a^b) = b*ln(a)

 

[Hacp]

3200 * (1.12)^n = 100000
(1.12)^n=100000/3200
So its log base 1.12(100000/3200)=n
n= 30.32

 

[OogyWaWa]

Divide and take the natural log of both sides. Are you sure this is finance homework and not high school algebra?

positive.
ok, so i've got:

31800 * (1.12)^n = 138750
1.12^n = 138750 / 31800
1.12^n = 4.36

then... ln(1.12) + ln(4.36) ???


the original problem:
Today your account was credited with its annual interest of $16,650 . The account was
established some time ago with a $31,800 initial deposit. No other deposits or
withdrawals have been made. The account earns 12.0% annual interest. How many
years ago was the account established?

when he did it in class he saved it in his calc so i couldn't see how he did the algebra...

 

[Hacp]

positive.
ok, so i've got:

31800 * (1.12)^n = 138750
1.12^n = 138750 / 31800
1.12^n = 4.36

then... ln(1.12) + ln(4.36) ???


the original problem:
Today your account was credited with its annual interest of $16,650 . The account was
established some time ago with a $31,800 initial deposit. No other deposits or
withdrawals have been made. The account earns 12.0% annual interest. How many
years ago was the account established?

when he did it in class he saved it in his calc so i couldn't see how he did the algebra...

I already explained it. Dont' take the natural log. Take the log base 1.2. Here is my calculator I used.
http://rechneronline.de/logarithm/

 

[OogyWaWa]

I already explained it.

can you explain how you get the log base of something? I need to know how to do it on a regular calculator...

nm:

ln(4.36)/ln(1.12) does it

thanks all

 

[Hacp]

The answer should be 14 with your calculations.
I don't really know what your professor did. Here is how I did your problem. Here is the calculator I used to do the logs. http://rechneronline.de/logarithm/

 

[Hacp]

can you explain how you get the log base of something? I need to know how to do it on a regular calculator...

nm:

ln(4.36)/ln(1.12) does it

thanks all

Yes you can do that too because of the rule that allows you to bring powers to the front.

 

[OogyWaWa]

The answer should be 14 with your calculations.
I don't really know what your professor did. Here is how I did your problem. Here is the calculator I used to do the logs. http://rechneronline.de/logarithm/

wow, thanks for the effort.

I ended up doing:

1.12^n = 4.36
ln(4.36)/ln(1.12) = n
n = 12.9929
n+1 (round) = 14

 

[OogyWaWa]

atot <3 off to do more finance @_@

 

[gaidensensei]

The answer should be 14 with your calculations.
I don't really know what your professor did. Here is how I did your problem. Here is the calculator I used to do the logs. http://rechneronline.de/logarithm/

Nice work, glad to see folks helping out another.