# Simplying Log Problems

• January 11th 2010, 11:20 PM
metallica007
Simplying Log Problems
Hi My Friends
I need ur help to simply the following problem
http://img20.imageshack.us/img20/4176/logt.jpg

can u please explain the steps to solve the problem

• January 12th 2010, 12:15 AM
Hello metallica007
Quote:

Originally Posted by metallica007
Hi My Friends
I need ur help to simply the following problem
http://img20.imageshack.us/img20/4176/logt.jpg

can u please explain the steps to solve the problem

The definition of a log is this (and it's a bit involved!):
The log of a number to a certain base is that power to which the base must be raised in order to get the number.
In other words, if
$x = \log_b(y)$
then the log of $y$ to base $b$ is $x$. So when we raise $b$ to the power $x$ we get $y$. In other words:
$b^x = y$
Make sure you can get your mind around that!

The laws of logs (whatever the base) are:
$\log (a) + \log (b) = \log(ab)$... (1)

$\log(a) -\log(b) = \log\left(\frac{a}{b}\right)$
... (2)

$a\log(b) = \log\left(b^a\right)$
... (3)
Now to your question. Note first that $\sqrt5 = 5^{\frac12}$. So we get:
$\log_5\left(\frac{\sqrt5}{5}\right) = \log_5(\sqrt5) - \log_5(5)$, using law (2) above
$=\log_5(5^{\frac12}) - \log_5(5^1)$

$=\tfrac12-1$, using the definition of a log

$=-\tfrac12$
• January 12th 2010, 08:12 AM
metallica007
Quote:

Hello metallica007The definition of a log is this (and it's a bit involved!):
The log of a number to a certain base is that power to which the base must be raised in order to get the number.
In other words, if
$x = \log_b(y)$
then the log of $y$ to base $b$ is $x$. So when we raise $b$ to the power $x$ we get $y$. In other words:
$b^x = y$
Make sure you can get your mind around that!

The laws of logs (whatever the base) are:
$\log (a) + \log (b) = \log(ab)$... (1)

$\log(a) -\log(b) = \log\left(\frac{a}{b}\right)$
... (2)

$a\log(b) = \log\left(b^a\right)$
... (3)
Now to your question. Note first that $\sqrt5 = 5^{\frac12}$. So we get:
$\log_5\left(\frac{\sqrt5}{5}\right) = \log_5(\sqrt5) - \log_5(5)$, using law (2) above
$=\log_5(5^{\frac12}) - \log_5(5^1)$

$=\tfrac12-1$, using the definition of a log

$=-\tfrac12$