1. ## Simplying Log Problems

Hi My Friends
I need ur help to simply the following problem

can u please explain the steps to solve the problem

2. Hello metallica007
Originally Posted by metallica007
Hi My Friends
I need ur help to simply the following problem

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
$\displaystyle x = \log_b(y)$
then the log of $\displaystyle y$ to base $\displaystyle b$ is $\displaystyle x$. So when we raise $\displaystyle b$ to the power $\displaystyle x$ we get $\displaystyle y$. In other words:
$\displaystyle b^x = y$
Make sure you can get your mind around that!

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

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

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

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

$\displaystyle =-\tfrac12$

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
$\displaystyle x = \log_b(y)$
then the log of $\displaystyle y$ to base $\displaystyle b$ is $\displaystyle x$. So when we raise $\displaystyle b$ to the power $\displaystyle x$ we get $\displaystyle y$. In other words:
$\displaystyle b^x = y$
Make sure you can get your mind around that!

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

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

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

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

$\displaystyle =-\tfrac12$