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Wize High School Grade 12 Pre-Calculus Textbook > Logarithmic Functions

Basics of Logarithms

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Basics of Logarithms

The inverse of an exponential function, y=bxy=b^x, can be expressed in two ways:

x=by→Exponential Form of the Inversey=log⁡bx→Logarithmic Form of the Inverse\begin{array}{rcl} x=b^y&\color{red}\rightarrow&\textbf{Exponential Form}~\text{of the Inverse}\\\\ y=\log_{b}x&\color{red}\rightarrow&\textbf{Logarithmic Form}~\text{of the Inverse} \end{array}

Wize Tip
The logarithm function is the inverse of the exponential function.

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Evaluating Logarithms

Since x=by  &  y=log⁡bxx=b^y~~\&~~y=\log_{b}x are equivalent, then evaluating logarithms is similar to evaluating exponents.

We can ask ourselves the question: "b\colorFive{\textbf{b}} raised to what power (y) results in x\colorFive{\textbf{x}}?"


Example 1
The value of log⁡28\log_{2}8 is 33 since 23=82^3=8.
So, 2 raised to the power of 3 results in 8.
Therefore, log⁡28=3\boxed{\log_{2}8=3}.

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Example 2

Approximate log⁡312\log_{3}12 to the nearest tenth.

Let's find two values of log⁡3x\log_{3}x close to log⁡312\log_{3}12.

Since 32=9  &  33=273^2=9~~\&~~3^3=27, then:

log⁡39<log⁡312<log⁡3272<log⁡312<3\begin{array}{rcccl} \log_{3}9&<&\log_{3}12&<&\log_{3}27\\\\ 2&<&\log_{3}12&<&3 \end{array}

12 is much closer to 9 than 27.

∴ log⁡312≈2.3\therefore~\log_{3}12\approx2.3

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Natural Logarithm

There are many different bases that can be used with logarithms, however, the most common bases are base 10 and base ee, where ee is an irrational number with an approximate value of 2.718282.71828.

The natural logarithm has base ee and is written as:
y=ln⁡x\boxed{y=\ln{x}}

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Example: Basics of Logarithms

Evaluate the following:
  1. log⁡464\log_{4}{64}
  2. log⁡33\log_{3}{\sqrt{3}}
  3. log⁡218\log_{2}{\dfrac{1}{8}}

Part a.

Let y=log⁡464y=\log_{4}64.

Then,
4y=644y=43→y=3∴ log⁡464=3\begin{array}{rcl} 4^{y}&=&64\\\\ 4^y&=&4^3&\color{red}\rightarrow&y&=&3\\\\ \therefore~\log_{4}{64}&=&3 \end{array}


Part b.

Let y=log⁡33y=\log_{3}{\sqrt{3}}.

Then,
3y=33y=312→y=12∴ log⁡33=12\begin{array}{rcl} 3^y&=&\sqrt{3}\\\\ 3^y&=&3^{\frac{1}{2}}&\color{red}\rightarrow&y&=&\dfrac{1}{2}\\\\ \therefore~\log_{3}{\sqrt{3}}&=&\dfrac{1}{2} \end{array}


Part c.

Let y=log⁡218y=\log_{2}{\dfrac{1}{8}}.

Then,
2y=182y=8−12y=2−3→y=−3∴ log⁡218=−3\begin{array}{rcl} 2^y&=&\dfrac{1}{8}\\\\ 2^y&=&8^{-1}\\\\ 2^y&=&2^{-3}&\color{red}\rightarrow&y&=&-3\\\\ \therefore~\log_{2}{\dfrac{1}{8}}&=&-3 \end{array}

Practice: Basics of Logarithms

Match each exponential expression to its inverse.
A.
127=3−3\dfrac{1}{27}=3^{-3}
B.
27=(13)−327={\Bigg(\dfrac{1}{3}\Bigg)}^{-3}
C.
256=44256=4^4
D.
y=bxy=b^{x}
E.
16=2416=2^4
F.
x=byx=b^y
4=log⁡2164=\log_{2}16
4=log⁡42564=\log_{4}256
−3=log⁡3(127)-3=\log_{3}{\Bigg(\dfrac{1}{27}\Bigg)}
x=log⁡byx=\log_{b}y
y=log⁡bxy=\log_{b}x
3=log⁡3273=\log_{3}27

Practice: Basics of Logarithms

Evaluate each logarithm and match it to its corresponding answer.

A.
log⁡255\log_{25}5
B.
log⁡2128\log_{2}\sqrt{128}
C.
log⁡927\log_{9}27
D.
log⁡432\log_{4}32
52\dfrac{5}{2}
32\dfrac{3}{2}
72\dfrac{7}{2}
12\dfrac{1}{2}

Practice: Basics of Logarithms

Evaluate log⁡18432\log_{\frac{1}{8}}{^4\sqrt{32}}.