# Math Help - Condensing Logarithms

1. ## Condensing Logarithms

Okay so I am totally confused on how to do this prroblem it asking to condense the following problem:
3log_4a+5log_4b-log_4c

2. Originally Posted by jvrobi167
Okay so I am totally confused on how to do this prroblem it asking to condense the following problem:
3log_4a+5log_4b-log_4c

Log questions are pre-calc not trig

Either way use the power law and the addition/subtraction laws.

Power Law: $k\log_b(a) = \log_b(a^k)$

Addition Law: $\log_b(a) + \log_b(c) = \log_b(ac)$

Subtraction Law: $\log_b(a) - \log_b(c) = \log_b \left(\frac{a}{c}\right)$

For example $\log_2(3) + 3\log_2(6) = \log_2(648)$

( $648 = 6^3 \times 2$ in case you were wondering)

3. Originally Posted by jvrobi167
Okay so I am totally confused on how to do this prroblem it asking to condense the following problem:
3log_4a+5log_4b-log_4c

To "condense" a log expression like this, use the following "laws of logarithms"...

$log_xy^k=klog_xy$

$log_xa+log_xb=log_x(ab)$

$log_xa-log_xb=log_x\left(\frac{a}{b}\right)$

Hence,

$3log_4a+5log_4b-log_4c=log_4a^3+log_4b^5-log_4c$

$=log_4\left(a^3b^5\right)-log_4c$

Try to finish using the law of subtraction.