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Exponential and Logarithmic Functions

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Algebra 2
Exponential and Logarithmic Functions
Evaluating Logarithms

Evaluating Logarithms

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Concept

Logarithm

A logarithm is the inverse of an exponential function. The logarithm of a positive number mm can be written as follows. The expression log⁡b(m)\log_{b}(m) is read as log⁡\log base bb of mm and states that raising bb to the nthn^{\text{th}} power yields m.m.

log⁡b(m)=n⇔bn=m\log_{b}(m)=n \quad \Leftrightarrow \quad b^n=m

Here, bb is both the base of the logarithm and the base of the exponent. For example, the value of nn in log⁡4(16)=n\log_{4}(16)=n is given by the exponent to which 44 would would be raised to result in 16.16. log⁡4(16)=n⇔4n=16log⁡4(16)=2⇔42=16\begin{aligned} \log_{4}(16)={\color{#0000FF}{n}}\quad & \Leftrightarrow\quad 4^{\color{#0000FF}{n}}=16\\ \log_{4}(16)={\color{#0000FF}{2}}\quad & \Leftrightarrow\quad 4^{\color{#0000FF}{2}}=16 \end{aligned} The logarithmic form and exponential form are equivalent.

The relationship between base and exponent for logarithms and powers
Concept

Common Logarithm

A common logarithm is a logarithm of base 10.10. For example, log⁡10(1000)\log_{10}(1000) is equal to 33 since 10310^3 is equal to 1000.1000.

the Connection between the base and the exponent for common logarithms

Since log⁡10\log_{10} is used so often, it is sometimes written without a base. For positive values of m,m, the common log⁡\log of mm can be defined as follows.

log⁡(m)=n⇔10n=m\log(m)=n \quad \Leftrightarrow \quad 10^n=m

Concept

The Natural Logarithm

A natural logarithm, written ln⁡,\ln, is a logarithm with base ee. This means that ln⁡(m)\ln(m) equals the exponent to which ee must be raised to equal m.m.

ln⁡(m)=n⇔m=en\ln(m)=n \quad \Leftrightarrow \quad m = e^n

Note that it is possible to write log⁡e\log_{e} instead of the more common ln⁡.\ln.

ln⁡(m)=log⁡e(m) \ln(m) = \log_{e}(m)
Exercise

For the following expressions, rewrite them in either logarithmic or exponential form. log⁡(100)=282=64log⁡2(32)=5e0=1 \log(100)=2 \quad 8^2=64 \quad \log_2(32)=5 \quad e^0=1

Solution

We will rewrite each expression one at a time using the relationship log⁡b(m)=n⇔bn=m. \log_{b}(m)=n \quad \Leftrightarrow \quad b^n=m. Notice that the first and third are written as logarithms and the second and fourth are written as exponents. Since a base is not written on the first logarithm, we know it's base 10.10. Furthermore, 22 is the exponent. Thus, log⁡(100)=2⇔102=100. \log(100)=2 \quad\Leftrightarrow\quad 10^2=100. The same reasoning applies for the third expression, log⁡2(32)=5.\log_{2}(32)=5. log⁡2(32)=5⇔25=32 \log_{2}(32)=5 \quad \Leftrightarrow \quad 2^5=32 Using the same relationship in the opposite way, we can rewrite the second and fourth expression. In 82=64,8^2=64, it can be seen that 88 is the base and 22 is the exponent. This means that 6464 is the value of which we take the logarithm. 82=64⇔log⁡8(64)=2. 8^2=64 \quad \Leftrightarrow \quad \log_{8}(64)=2. For the last expression, this yields e0=1⇔log⁡e(1)=0⇔ln⁡(1)=0. e^0=1 \quad \Leftrightarrow \quad \log_{e}(1)=0 \quad \Leftrightarrow \quad \ln(1)=0. To summarize, the following expressions are equivalent. log⁡(100)=2⇔102=10082=64⇔log⁡8(64)=2log⁡2(32)=5⇔25=32e0=1⇔ln⁡(1)=0\begin{aligned} \log(100)=2 &\quad\Leftrightarrow\quad 10^2=100\\ 8^2=64 &\quad\Leftrightarrow\quad \log_8(64)=2\\ \log_2(32)=5 &\quad\Leftrightarrow\quad 2^5=32\\ e^0=1 &\quad\Leftrightarrow\quad \ln(1)=0 \end{aligned}

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Exercise

Evaluate the following logarithms. log⁡3(81)ln⁡(e)log⁡(11000) \log_{3}(81) \quad \ln(e) \quad \log \left(\frac{1}{1000} \right)

Solution

When evaluating logarithms, it can be helpful to think about what the expression means. log⁡3(81)\log_{3}(81) asks the exponent to which 33 must be raised to equal 81.81. Since, 3⋅3⋅3⋅3=81,3 \cdot 3 \cdot 3 \cdot 3 =81, log⁡3(81)=4. \log_{3}(81)=4. The second expression, ln⁡(e),\ln(e), is a logarithm of base e.e. It asks the exponent to which ee must be raised to equal e.e. Since e1=e,e^1=e, ln⁡(e)=1. \ln(e)=1. Notice that, unlike the other expressions, the last, which is of base 10,10, contains a fraction. Thus, we must consider the exponent to which 1010 is raised to equal 11000.\frac{1}{1000}. As it turns out, negative exponents yield fractions. x-n=1xn. x^{\text{-} n} = \dfrac{1}{x^n}. It can be helpful to rewrite 10001000 as a power of 10.10. Since 1000=103,1000 = 10^3, 11000=1103=10-3. \frac{1}{1000} = \frac{1}{10^3} = 10^{\text{-} 3}. Thus, log⁡(11000)=-3.\log \left(\frac{1}{1000} \right)=\text{-} 3.

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Exercises

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