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
please help
Log questions are pre-calc not trig
Either way use the power law and the addition/subtraction laws.
Power Law: $\displaystyle k\log_b(a) = \log_b(a^k)$
Addition Law: $\displaystyle \log_b(a) + \log_b(c) = \log_b(ac)$
Subtraction Law: $\displaystyle \log_b(a) - \log_b(c) = \log_b \left(\frac{a}{c}\right)$
For example $\displaystyle \log_2(3) + 3\log_2(6) = \log_2(648)$
($\displaystyle 648 = 6^3 \times 2$ in case you were wondering)
To "condense" a log expression like this, use the following "laws of logarithms"...
$\displaystyle log_xy^k=klog_xy$
$\displaystyle log_xa+log_xb=log_x(ab)$
$\displaystyle log_xa-log_xb=log_x\left(\frac{a}{b}\right)$
Hence,
$\displaystyle 3log_4a+5log_4b-log_4c=log_4a^3+log_4b^5-log_4c$
$\displaystyle =log_4\left(a^3b^5\right)-log_4c$
Try to finish using the law of subtraction.