# working with logarithms?

• Oct 7th 2010, 05:16 PM
yolanda
working with logarithms?
so my advanced functions teacher didn't really explain this lesson well and i can't seem to figure it out on my own.

http://www4a.wolframalpha.com/Calcul...=36&w=119&h=37

if you could be kind enough to explain the process i would greatly appreciate it!
• Oct 7th 2010, 05:57 PM
harish21
The logarithm of a number y with respect to a base b is the exponent to which we have to raise b to obtain y.

the definition would be:

$x = log_{b} y \Longleftrightarrow b^x = y$

and we say that x is the logarithm of y with base b if and only if b to the power x equals y.

For example:

$10^2 = 100 \Longleftrightarrow log_{10} 100 = 2$

and, $10^{-2} = 0.01 \Longleftrightarrow log_{10} 0.01 = -2$

Based on this, can you now complete your question?!
• Oct 7th 2010, 06:13 PM
Quote:

Originally Posted by yolanda
so my advanced functions teacher didn't really explain this lesson well and i can't seem to figure it out on my own.

http://www4a.wolframalpha.com/Calcul...=36&w=119&h=37

if you could be kind enough to explain the process i would greatly appreciate it!

$\displaystyle\ \log_9\left(\frac{1}{3}\right)=x$

$\displaystyle\ \log_3\left(\frac{1}{9}\right)=y$

Imagine a "fog" surrounding the "log" and the base slides over to the other side,
so we get

$\displaystyle\frac{1}{3}=9^x$

$\displaystyle\frac{1}{9}=3^y$

Now x and y will be easier to derive.
• Oct 11th 2010, 09:30 AM
yolanda
thank you