# Logs

• Nov 11th 2005, 08:16 PM
kurtolios
How do I solve this??
20,000 = 10,000 (1+.05)^n

n is an exponent
I need to learn about logs and how do use the log button on my calculator.. I think someone was helping me once and they used a 1/x button or a y^x button on my calculator... any ideas would be greatly appreciated
• Nov 12th 2005, 05:04 AM
hpe
Quote:

Originally Posted by kurtolios
20,000 = 10,000 (1+.05)^n

n is an exponent
I need to learn about logs and how do use the log button on my calculator.. I think someone was helping me once and they used a 1/x button or a y^x button on my calculator... any ideas would be greatly appreciated

First divide both sides of the equation by 10,000 to simplify the question. A trial and error approach works as follows: compute 1.05^1, 1.05^2, 1.05^3 and so on, until you get close to the answer. Then continue with fractions as exponents, e.g. 1.05^4.5, 1.05^4.7 and so on. On the TI-83, this is done by keying in "1.05 y^x 4.7", for example (I believe).

You might get sick and tired of this, so then look up logarithms in your textbook, and find an exercise or example that sort of looks like this problem. Try to understand how it was done, and you will be on your way.
• Nov 12th 2005, 05:58 AM
Jameson
Quote:

Originally Posted by hpe
First divide both sides of the equation by 10,000 to simplify the question. A trial and error approach works as follows: compute 1.05^1, 1.05^2, 1.05^3 and so on, until you get close to the answer. Then continue with fractions as exponents, e.g. 1.05^4.5, 1.05^4.7 and so on. On the TI-83, this is done by keying in "1.05 y^x 4.7", for example (I believe).

You might get sick and tired of this, so then look up logarithms in your textbook, and find an exercise or example that sort of looks like this problem. Try to understand how it was done, and you will be on your way.

Trial and error? This can be done very simply using logs. I agree that first you should divide out the 10,000 to make the equation simpler. Once you are there, take note of these two important rules.

If a=b, then log(a)=log(b). A and B can be more than one term! You must take the log of everything on the left in order for it to equal the log of everything on the right.

And here's a more elegant rule.

\$\displaystyle \log{(a^n)}=n\log(a)\$

If you work it out right, you'll get an exact answer. Off you go.
• Nov 12th 2005, 06:30 AM
hpe
Quote:

Originally Posted by Jameson
Trial and error? This can be done very simply using logs.

Of course you're right. Some drudgery with the calculator can be excellent motivation for further study, though. ;)