The rules of logarithmic and exponential functions and the properties of these functions are presented. The use of these rules to simplify expressions and solve equations are explained through examples and questions with detailed solutions.
In what follows we use the following notations:
log_{b} represents the logarithm of base b
ln represents the logarithm of base e
Rules of the Logarithmic functions
For U and V real positive numbers, b positive and not equal to 1:

log_{b} 1 = 0
Example: log_{5} 1 = 0

log_{b} (U V) = log_{b} U + log_{b} V
Example: log_{2} (4 × 8) = log_{2} 4 + log_{2} 8

log_{b} (U / V) = log_{b} U  log_{b} V
Example: log_{2} (8 / 4) = log_{2} 8  log_{2} 4

log_{b} (U ^{r}) = r log_{b} U
Example: log_{3} (9 ^{2}) = 2 log_{3} 9

log_{b} (b ^{x}) = x
Example: log _{5} (5 ^{2}) = 2

b ^{ (log b x) } = x , x > 0
Example: 2 ^{(log 2 5)} = 5

log_{b} U = log_{b} V ⇔ U = V
Example: ln (x + 1) = ln ( x + 4) ⇔ x + 1 =  x + 4
Example 1
Use the rules of the Logarithmic functions to simplify the following expressions
a) \( \log_9 1\) , b) \( \log_2 (4 \times 16) \) , c) \( \log_3 \dfrac{3}{27}\) , d) \( \log_4 4^6\) ,
e) \( \log_2 2^{(x+1)} \) , f) \( e^{\ln(x^2+1)} \)
Solution to Example 1
a) \( \log_9 1 = 0\) , use rule 1.
b) \( \log_2 (4 \times 16) = \log_2 4 + \log_2 16 = \log_2 2^2 + \log_2 2^4 = 2 + 4 = 6\) , use rules 2 and 5.
c) \( \log_3 \dfrac{3}{27} = \log_3 3  \log_3 27 = \log_3 3^1  \log_3 3^3 = 1  3 = 2 \) , use rules 3 and 5.
d) \( \log_4 4^6 = 6\) , use rule 5.
e) \( \log_2 2^{(x+1)} = x + 1\) , use rule 6.
f) \( e^{\ln(x^2+1)} = x^2+1 \) , use rule 5.
Rules of the Exponential functions
For \(x\) and \( y\) real numbers and \( a \) real and positive:
 \( a^n = a \times a \times a .... \times a \quad ( \, n \, \text{times}) \)
Example: \( 2^4 = 2 \times 2 \times 2 \times 2 = 16 \)
 \( a^0 = 1 \)
Example: \( 100^0 = 1 \)
 \( a^{x} = \dfrac{1}{a^x} \)
Example: \( 6^{2} = \dfrac{1}{6^2} \)
 \( a^x a^y = a^{x+y} \)
Example: \( 4^2 4^3 = 4^{2+3} = 4^5 \)
 \( \dfrac{a^x}{a^y} = a^{xy} \)
Example: \( \dfrac{5^6}{5^4} = 5^{64} = 5^{2} \)
 \( (a^x)^y = a^{xy} \)
Example: \( (7^2)^3 = 7^{2\times 3} = 7^6\)
Example 2
Use the rules of the exponential functions to simplify the following expressions
a) \( 2^3 2^{3} \) , b) \( \dfrac{4^3 4^x}{4^{2x}} \) , c) \( {(4^2)}^{2x} \)
Solution to Example 2
a) \( 2^3 2^{3} = 2^{33} = 3^0 = 1 \) , use rules 4 and 2.
b) \( \dfrac{4^3 4^x}{4^{2x}} = \dfrac{4^{3+x}}{4^{2x}} = 4^{3+x2x} = 4^{3x} \) , use rules 4 and 5.
c) \( {(4^2)}^{2x} = 4^{ 2 \times 2x } = 4^{4x} = (4^4)^{x} = 256^x \) , use rule 6
Inverse Rule  Relationship Between the Logarithmic and Exponential Functions
The exponential and logarithmic functions are inverses of each other. Hence
\[ \log_b x = y \iff x = b^y \]
Notes
a  The bases of the logarithmic and exponential functions are the same.
b  \( y \) is the exponent in the exponential form which means \( \log_b x \) is the exponent the base \( b \) must be raised to obtain \( x \).
Example 3
Use the inverse rule to write each of the following in the form of a logarithm.
a) \( 2^4 = 16 \) , b) \( 3^3 = 27 \), c) \( e^5 = y\)
Solution to Example 3
a) \( 2^4 = 16 \) is equivalent to \( \log_2 16 = 4 \), since 4 is the exponent of 2 that gives 16.
b) \( 3^3 = 27 \) is equivalent to \( \log_3 27 = 3 \), since 3 is the exponent of 3 that gives 27.
b) \( e^5 = y\) is equivalent to \( \ln y = 5 \), since 5 is the exponent of e that gives 25.
Example 4
Use the inverse rule to write each of the following in the form of an exponential.
a) \( \log_b 27 = x \) , b) \( \log_{36} 6 = 1 / 2 \) , c) \( \log_2 (1 / 8) = 3 \) , d) \( \log_8 2 = 1 / 3\)
Solution to Example 4
a) The logarithmic form \( \log_b 27 = x \) is equivalent to the exponential form \( 27 = b^x \)
b) The logarithmic form \( \log_{36} 6 = 1 / 2 \) is equivalent to the exponential form \( 6 = 36^{1/2} \)
c) \( \log_2 (1 / 8) = 3 \) is equivalent to \(1 / 8 = 2^{3}\)
d) \( \log_8 2 = 1 / 3\) is equivalent to \(2 = 8^{1/3}\)
Example 5
Use the inverse rule to solve the equation \( \log_3 x = 4\).
Solution to Example 5
Rewrite the given logarithmic equation in exponential form.
\( x = 3^4 = 81 \)
Example 6
Solve for x the following equations.
a) \( \log_3 x = 5 \)
b) \( \log_2 (x  3) = 2 \)
c) \( 2 \log_3 ( x + 1) = 6 \)
Solution to Example 6
a) To solve the equation \( \log_3 x = 5\), rewrite it into exponential form
\( x = 3^5 \)
b) Rewrite the equation \( \log_2 (x  3) = 2 \) into exponential form
\( x  3 = 2^2 = 4 \)
Solve for x
x = 4 + 3 = 7
c) Divide all terms of the equation \( 2 \log_3 ( x + 1) = 6 \) by 2
\( \log_3 ( x + 1) = 3 \)
Rewrite the equation obtained in exponential form
\(  x + 1 = 3^3 = 27 \)
Solve for x
x =  26
Questions with Solutions
Part 1

Write as one logarithm.
 \( 2 \log_a x + \log_a (x+1) \)
 \( 2 \log_2 (x+2) \log_2(x1) \)

Change each exponential expression to logarithmic expression.
 \( 3^4 = 81\)
 \( 4^{1/2} = 2 \)
 \( 3^{3} = 1 / 27 \)
 \( 10^3 = 1000\)

Change each logarithmic expression to an exponential expression and solve for x.
 \( \log_b x = 2 \)
 \( 2 \log_2 (x+2) = 6 \)

Write as one logarithm and solve for x.
 \( \log_a 2 + \log_a (x  1) = 3 \)
 \( \log_2 (x+2)  \log_2(x1) = 2 \)
Part 2
 Find parameters a and b so that f(0) = 0 and f(1) = 2, where f is a logarithmic function given by
\[ f(x) = a \log_2(x + b) \]
 Find the x intercept of the graph of f where f is a function given by
\[ f(x) = \log_2(2 x + 6) \]
Solutions to the Above Questions
Part 1

Write as one logarithm.
 \( 2 \log_a x + \log_a (x+1) = \log_a x^2 + \log_a (x+1) = \log_a x^2(x+1)\)
 \( 2 \log_2 (x+2) \log_2(x1) = \log_2 (x+2)^2  \log_2(x1) = \dfrac{(x+2)^2}{x1} \)

Change each exponential expression to logarithmic expression.
 The exponential form \( 3^4 = 81 \) is equivalent to the logarithmic form
\( 4 = \log_3 81 \)
 The exponential form \( 4^{1/2} = 2 \) is equivalent to the logarithmic form
\( 1 / 2 = \log_4 2 \)
 \( 3^{3}= 1 / 27 \) in logarithmic form is given by
\( 3 =\log_3 (1/ 27) \)
 \( 10^3 = 1000\) in logarithmic form is given by
\( 3 = \log_{10} 1000\)

Change each logarithmic expression to an exponential expression and solve for x.
 \(\log_b x = 2 \iff x = b^2\)
 \( 2\log_2 (x+2) = 6 \)
Divide both sides by 2 and rewrite as
\(\log_2 (x+2) = 3\)
Rewrite in exponential form
\( x + 2 = 2^3\)
Solve for x
\( x = 6 \)

Write as one logarithm and solve for x.
 \(\log_a 2 +\log_a (x  1) = 3 \)
Use rule 2 of logarithms to group the logarithms on the left and write the above equation as
\(\log_a 2 (x  1) = 3 \)
Use the inverse rule to rewrite as
\( 2(x  1) = a^3\)
Solve for x
\( x = \dfrac{a^3}{2} +1\)
 \(\log_2 (x+2)\log_2(x1) = 2 \)
Use rule 3 of logarithms to group the logarithms on the left and write the above equation as
\(\log_2 \dfrac{x+2}{x1} = 2 \)
Use the inverse rule to rewrite as
\( \dfrac{x+2}{x1} = 2^{2}\)
Solve for x
\( \dfrac{x+2}{x1} = 4\)
\( x+2 = 4(x  1) \)
\( x = 2 \)
Part 2

Use the fact that \( f(0) = 0\) to obtain the equation
\( a \log_2(0 + b) = 0 \)
Divide both sides by \( a \) to obtain
\( \log_2 b = 0 \)
Rewrite the above using exponential form to solve for b
\( b = 2^0 = 1\)
Function f can be written as
\( f(x) = a \log_2(x + 1) \)
Next, use the fact that \( f(1) = 2 \) to obtain the equation
\( a \log_2(1 + 1) = 2 \)
Simplify to obtain
\( a = 2 \)
Function f is given by
\( f(x) = 2 \log_2(x + 1) \)

The x intercept is found in solving \( \log_2(2 x + 6) = 0 \)
Rewrite the above in exponential form
\( 2x + 6 = 2^0 = 1 \)
Solve the above equation to obtain the x intercept at
\( x =  5/2 \)
More References and Links to logarithmic FunctionsLogarithmic Functions (interactive tutorial).
Solve Logarithmic Equations  Detailed Solutions
How Solve Exponential Equations Questions with Detailed Solutions
Graph Logarithmic Functions.
Tutorial on Exponential and Logarithmic Equations.
Graph of Logarithmic and Exponential Functions  Self Test.
Solve Exponential and Logarithmic Equations (self test). 