Digging Laws of Logarithm Deeper

In order to evaluate logarithmic expression we need to know the basic laws of logarithm. This topic is still not that easy because it’s very confusing. On how to deal with this easily, practice is the key. This website is dedicated to that.

Basic Laws:

I. Addition-Multiplication Law

\log (xy)=\log(x)+\log(y)

Example:

\log(42)=\log (7\cdot 6)= \log7+\log6

Think of this,

\log(7)\cdot \log(6)\ne \log(7\cdot 6) \ne \log(42) \log(2+3)\ne \log(2\cdot 3)

This is the common misconception about this addition law.

 

II. Subtraction-Division Law

\log(\displaystyle\frac{x}{y})=\log(x)-\log(y)

Example:

\log(\frac{2}{3})=\log(2)-\log(3)

Think of this,

\displaystyle\frac{\log(2)}{\log(3)}\ne \log(2)-\log(3)

 

III. Power Law

\log(u^n)=n\log(u)

Example:

\log(25)=\log(5^2)=2\log(5) \log(x^2+2x+1)=\log(x+1)^2=2\log(x+1)

 

Worked Problem 1:

Given \log(3)=p, \log(2)=q, \log(7)=r. Find the value $latex

\log(84)$

 

[toggle title=”Solution”]

We need to express \log(84) in terms of the given quantities.

$latex \log(84)=\log(4\cdot 3\cdot 7)= \log(2^2\cdot 3\cdot 7)=\log(2^2)+\log(3)+\log(7)

\to 2\log(2)+\log(3)+\log(7)=2q+p+r$ [/toggle]

 

Worked Problem 2: 14th Philippine Math Olympiad Qualifying Round

Let r=\log 50 and s=\log 80. Express 7\log 20 in terms of r and s.

(a)2r+s                  (b) 2r+3s              (c) r+2s                 (d) 3r+2s

 

[toggle title=”Solution”]

Using power rule,

7\log 20=\log 20^7=\log(20)(20)(20)(20)(20)(20)(20) Work first if how many \log 80 we can get out of \log(20)(20)(20)(20)(20)(20)(20)

\log(20)(20)(20)(20)(20)(20)(20)=\log(80^3\cdot 50^2)=\log(80^3)+\log(50^2)=3\log80+2log50=3s+2r. The answer is letter (b). [/toggle]

 

Worked Problem 3: 2007 MMC Elimination

Evaluate and simplify: \displaystyle\frac{\log_2\sqrt{8}}{log_2\root 3\of{4}}

 

[toggle title=”Solution”]

Take note that \log_bb^n=n.

The easiest approach to this problem is to express \sqrt{8} and \root 3\of{4} to the base 2 since the base of logarithm is 2.

\sqrt{8}=8^{\frac{1}{2}}=(2^3)^{\frac{1}{2}}=2^{\frac{3}{2}} \root 3\of{4}=4^{\frac{1}{3}}=(2^2)^{\frac{1}{3}}=2^{\frac{2}{3}}

 

Going back to the original expression,

\displaystyle\frac{\log_2\sqrt{8}}{log_2\root 3\of{4}} \displaystyle\frac{\log_22^{\frac{3}{2}}}{log_22^{\frac{2}{3}}} \displaystyle\frac{\frac{3}{2}}{\frac{2}{3}}

\displaystyle\frac{9}{4} [/toggle]

 

Worked Problem 4: 2013 MMC Elimination

If p=\log2, q=\log5 and r=\log7, express \log50+2\log70-\log_27 in terms of p,q, and r.

 

[toggle title=”Solution”]

Take note for this identity taken from change of base.

\log_ba=\displaystyle\frac{\log_c(a)}{\log_c(b)} take note of this identity and we will use this later.

$latex \log50+2\log70-\log_27

=\log(5^2\cdot2)+2\log(2\cdot 5\cdot 7)-\log_27

=2\log5+\log2+2(\log2+\log5+\log7)-\displaystyle\frac{\log7}{\log2}

=2\log5+\log2+2\log2+2\log5+2\log7)-\displaystyle\frac{\log7}{\log2}

=4\log5+3\log2+2\log7-\displaystyle\frac{\log7}{\log2}

=4q+3p+2r-\displaystyle\frac{r}{p}$ [/toggle]

 

Don’t miss the next topic about change of base property of logarithm.

 

 

 

 

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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