Logarithms

**Sxtxs** · April 6th 2010, 07:22 AM

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

**Prove It** · April 6th 2010, 07:28 AM

Originally Posted by Sxtxs

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

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

$= 4\log_a{2}$ .

Therefore $x = 4\log_a{2}$

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

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

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

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

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

So $\log_a{3} = y - \frac{x}{2}$ .

**integral** · April 6th 2010, 04:46 PM

Suppose:
$\textrm{log}_a16=x$
$\textrm{log}_a12=y$

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

We can know that:
$<br /> a=\sqrt[y]{12}$ or
$a=\sqrt[x]{16}$

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

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

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

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

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

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

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

Therefore:

$y=log_a12$

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

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

$\delta=xlog_{16}3$

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

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

$x=4log_a2$

Test:

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

&
x=12log_32
So

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

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

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