# Laws of logarithms

• Mar 17th 2009, 11:42 AM
db5vry
Laws of logarithms
I am familiar with the laws of logarithms, and there are three exam questions here that are each worth three marks:

1] Show that if x > 0,
$\displaystyle \log_a x^k = k \log_a x$

2] Given that x > 0, y > 0, show that
$\displaystyle \log_a \frac{x}{y} = \log_a x - \log_a y$

3] Given that x > 0, y > 0, show that
$\displaystyle \log_a (xy) = \log_a x + \log_a y$

How would you prove these laws? Would you show how you reach them or possibly use an example to do this?
• Mar 17th 2009, 12:49 PM
Laws of Logarithms
Hello db5vry
Quote:

Originally Posted by db5vry
I am familiar with the laws of logarithms, and there are three exam questions here that are each worth three marks:

1] Show that if x > 0,
$\displaystyle \log_a x^k = k \log_a x$

2] Given that x > 0, y > 0, show that
$\displaystyle \log_a \frac{x}{y} = \log_a x - \log_a y$

3] Given that x > 0, y > 0, show that
$\displaystyle \log_a (xy) = \log_a x + \log_a y$

How would you prove these laws? Would you show how you reach them or possibly use an example to do this?

I think that you are expected to prove each of these by using the definition of a logarithm, which is:

'The log of a number to a given base is that power to which the base must be raised to give the number'

In other words, $\displaystyle l = \log_a x \iff a^l = x$

So you'd do number 1, for instance, like this:

$\displaystyle l = \log_ax$

$\displaystyle \Rightarrow a^l = x$

$\displaystyle \Rightarrow (a^l)^k = x^k$

$\displaystyle \Rightarrow a^{lk} = x^k$

$\displaystyle \Rightarrow \log_ax^k = lk = k\log_ax$

Do you want to have a go at the others in the same way?