# Logarithms

• Apr 6th 2010, 07:22 AM
Sxtxs
Logarithms
Given that log a 16= x and log a 12 = y, express loga3 in terms of x and y.
• Apr 6th 2010, 07:28 AM
Prove It
Quote:

Originally Posted by Sxtxs
Given that log a 16= x and log a 12 = y, express loga3 in terms of x and y.

$\displaystyle \log_a{16} = \log_a{(2^4)}$

$\displaystyle = 4\log_a{2}$.

Therefore $\displaystyle x = 4\log_a{2}$

$\displaystyle 2\log_a{2} = \frac{x}{2}$.

$\displaystyle \log_a{12} = \log_a{(3\cdot 4)}$

$\displaystyle = \log_a{3} + \log_a{2^2}$.

Therefore $\displaystyle y = \log_a{3} + 2\log_a{2}$

$\displaystyle y = \log_a{3} + \frac{x}{2}$.

So $\displaystyle \log_a{3} = y - \frac{x}{2}$.
• Apr 6th 2010, 04:46 PM
integral
Suppose:
$\displaystyle \textrm{log}_a16=x$
$\displaystyle \textrm{log}_a12=y$

Express $\displaystyle \textrm{log}_a3$ in terms of x&y.

We can know that:
$\displaystyle a=\sqrt[y]{12}$ or
$\displaystyle a=\sqrt[x]{16}$

We can also infer, if $\displaystyle log_a3=\delta$ that:

$\displaystyle a=\sqrt[\delta]{3}$

So we have:
$\displaystyle \sqrt[\delta]{3}=\sqrt[x]{16}=\sqrt[y]{12}$ and

$\displaystyle \sqrt[\delta]{3}=\sqrt[y]{12}\,\,\,\,\,\,\therefore$

$\displaystyle \delta=\textrm{log}_{12^{\frac{1}{y}}}3\,\,\,\,\,\ therefore$
$\displaystyle \delta=y\textrm{log}_{12}3\,\,\,\,\because log_{x^n}y=\frac{log_xy}{n}$

$\displaystyle y=\frac{\delta}{log_{12}3}$

$\displaystyle y=\frac{log_a3}{log_{12}3}$

Therefore:

$\displaystyle y=log_a12$

Like wise:
$\displaystyle \sqrt[\delta]{3}=\sqrt[x]{16}$

$\displaystyle \delta=log_{16^{\frac{1}{x}}}{3}$

$\displaystyle \delta=xlog_{16}3$

$\displaystyle x=\frac{\delta}{log_{16}3}$

$\displaystyle x=\frac{log_a3}{log_{16}3}$

$\displaystyle x=4log_a2$

Test:

If
$\displaystyle \delta=3$
then
$\displaystyle y=3log_312$
so
$\displaystyle \frac{3}{log_{12}3}=3log_312$

&
x=12log_32
So

$\displaystyle x=\frac{3}{log_{16}3}=12log_32$

The test values where derived from:
$\displaystyle \sqrt[\delta]{3}=\sqrt[x]{16}=\sqrt[y]{12}$