Condensing Logarithms

May 2010
1
0
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(Crying)
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
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(Crying)
Log questions are pre-calc not trig (Hi)


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)
 
Dec 2009
3,120
1,342
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(Crying)
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.
 
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