# Condensing Logarithms

#### 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

#### e^(i*pi)

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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 (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)

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"...

$$\displaystyle log_xy^k=klog_xy$$

$$\displaystyle log_xa+log_xb=log_x(ab)$$

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

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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