Properties of Logarithms

Definition

The following two equations are equivalent: by=xandy=logbx. b^y = x \qquad \text{and} \qquad y = \log_b{x}. Here, yy is the logarithm of xx to the base bb. In other words, bb needs to be raised to the power of yy to equal xx.

Logarithms are only defined when bb is positive and not equal to 11, and xx is positive.

Note that blogbx=x and logbby=y. b^{\log_b{x}} = x \text{ and } \log_b{b^y} = y. Two of the most common log bases you will encounter are log10\log_{10} and loge\log_{e} (ee is Euler's number). They are sometimes written as simply lg\lg and ln\ln, respectively.

Technique

Several important identities are used to relate logarithms to one another. They are known as logarithmic laws.

Products: logb(xy)=logb(x)+logb(y)Ratios: logb(xy)=logb(x)logb(y)Powers: logb(xp)=plogb(x)Roots: logb(xp)=logb(x)pChange of bases: logb(x)=loga(x)loga(b) \begin{array}{lrcl} \textbf{Products: } & \log_b{(xy)} &=& \log_b{(x)} + \log_b{(y)} \\ \\ \textbf{Ratios: } & \log_b{ \left(\frac{x}{y}\right) } &=& \log_b{(x)} - \log_b{(y)} \\ \\ \textbf{Powers: } & \log_b{(x^p)} &=& p\,\log_b{(x)} \\ \\ \textbf{Roots: } & \log_b{\left(\sqrt[p]x \right )} &=& \frac{\log_b{(x)}}{p} \\ \\ \textbf{Change of bases: } & \log_b{(x)} &=& \frac{\log_a{(x)}}{\log_a{(b)}} \end{array}

Here are a couple of example problems:

Evaluate log108+log105+log1025 \log_{10}{8} + \log_{10}{5} + \log_{10}{25}

log108+log105+log1025=log10(8525)=log101000=3 \begin{aligned} \log_{10}{8} + \log_{10}{5} + \log_{10}{25} &= \log_{10}{\left( 8 \cdot 5 \cdot 25 \right)} \\ &= \log_{10}{1000} \\ &= 3 \, _\square \end{aligned}

 

Solve for xx: logx=1log(x+3) \log{x} = 1 - \log{\left(x+3\right)}

Use the laws of logarithms to put xx into a single logarithm.
logx+log(x+3)=1log(x(x+3))=1 \log{x} + \log{(x+3)} = 1 \\ \log{\left ( x(x+3) \right )} = 1 This logarithmic expression can then be rewritten as an exponential.
x(x+3)=101 x(x+3) = 10^1 This gives us the quadratic equation x2+3x10=0 x^2 + 3x - 10 = 0 , which has the roots x=2 x = 2 and x=5 x = -5 .

We immediately have to reject x=5 x = -5 , because log(5) \log{(-5)} and log(5+3) \log{(-5+3)} are negative and, therefore, undefined. Thus, the correct answer is x=2. x = 2. \, _\square

Application and Extensions

Here is an application that uses all five logarithmic laws:

Rewrite log2x+6log8ylog4z \log_2{x} + 6\,\log_8{y} - \log_4{z} as a single logarithm:

log2x+6log8ylog4z=log2x+6log2ylog28log2zlog24=log2x+6log2y3log2z2=log2x+2log2ylog2z2=log2x+log2y2log2z=log2(xy2z) \begin{aligned} \log_2{x} + 6\,\log_8{y} - \log_4{z} &= \log_2{x} + 6\frac{\log_2{y}}{\log_2{8}} - \frac{\log_2{z}}{\log_2{4}} \\ &= \log_2{x} + \frac{6\,\log_2{y}}{3} - \frac{\log_2{z}}{2} \\ &= \log_2{x} + 2\,\log_2{y} - \frac{\log_2{z}}{2} \\ &= \log_2{x} + \log_2{y^2} - \log_2{\sqrt{z}} \\ &= \log_2{\left( \frac{xy^2}{\sqrt{z}} \right )} _\square \end{aligned}

Note by Arron Kau
5 years, 4 months ago

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