# write to single log

• Dec 5th 2008, 11:55 AM
rj2001
write to single log
How do I start this? to make it one log?
• Dec 5th 2008, 12:15 PM
terr13
$\displaystyle Log(a/b) = log(a) - log(b)$
$\displaystyle log(ab) = log(a) + log(b)$
$\displaystyle a log(b) = log(b^a)$
• Dec 5th 2008, 12:15 PM
mr fantastic
Quote:

Originally Posted by rj2001
How do I start this? to make it one log?

The start:

$\displaystyle 12 \log_6 \sqrt{5x - 4} - \log_6 \left(\frac{6}{x} \right) + \log_6 6$

$\displaystyle = 12 \log_6 (5x - 4)^{1/2} - \log_6 \left(\frac{6}{x} \right) + \log_6 6$

$\displaystyle = \log_6 (5x - 4)^6 - \log_6 \left(\frac{6}{x} \right) + \log_6 6$

where the first term is got using one of the usual rules.

Now add and subtract logs of the same base using the usual rules. Simplify.
• Dec 5th 2008, 12:19 PM
Soroban
Hello, rj2001!

Quote:

Write as one log: .$\displaystyle 12\log_6\sqrt{5x-4} - \log_6\left(\frac{6}{x}\right) + \log_6(6)$

$\displaystyle \log_6\left(\sqrt{5x-4}\right)^{12} - \bigg[\log_6(6) - \log_6(x)\bigg] + \log_6(6)$

.$\displaystyle = \;\log_6(5x-4)^6 - \log_6(6) + \log_6(x) + \log_6(6)$

. . $\displaystyle = \;\log_6(5x-4)^6 + \log_6(x)$

. . $\displaystyle = \;\log_6\bigg[x(5x-4)^6\bigg]$