# Algebra: Logarithmic and Exponential Functions

1.
$$lnx=3$$
Solution
2.
$$2\ln{(3x)}=4$$
Solution
3.
$$\ln{(x-2)}+\ln{(2x-3)}=2lnx$$
Solution
4.
$$\frac{\log{{(3x-7)}^{2}}}{\log{4}}=10$$
Solution
5.
$${e}^{x}=72$$
Solution
6.
$${e}^{x}+5=60$$
Solution
7.
$${4e}^{2x}=5$$
Solution
8.
$${2}^{x}=10$$
Solution
9.
$$100({14}^{2x})+6=10$$
Solution

# Logarithmic And Exponential Functions - Introduction

Exponential and logarithmic equations might seem daunting at first glance. Let’s break down exactly what they are and learn how to solve them!

# What Are Exponents and Logarithms?

An exponent is used to show that you’re multiplying something by itself a certain number of times. So,
$${2}^{3}$$
is a quicker way to write
$$2\times 2\times 2$$
— and both give the same answer:
$$8$$
.
Logarithms are the opposite of exponents. Consider this: We know that
$${2}^{3}=8$$
, but let's represent this equation in another way. Using logarithms, we can re-write this equation as
$$\log_{2}{8}=3$$
. The subscript
$$2$$
here is the base of the logarithm. This is similar to asking, "
$$2$$
raised to which power equals
$$8$$
?" The answer would be
$$3$$
.

# Useful Logarithm Rules

There are two logarithm rules often used to help solve equations: The natural log rule and the common logarithm rule.
The Natural Log Rule states that if
$${e}^{y}=x$$
, then
$$\ln{x}=y$$
, where
$$e$$
is a transcendental number much like
$$\pi$$
, and is 2.718281828459 approximately.
The Common Logarithm Rule states that if
$${b}^{a}=x$$
, then
$$\log_{b}{x}=a$$
.

# Exponent and Logarithm Examples

Let’s look at a couple equations and see these rules in action.
1. Given the equation
$$\ln{x}=3$$
, solve for
$$x$$
.
Since we know that
$$\ln{x}=y$$
is equivalent to
$${e}^{y}=x$$
, we can apply the transformation to this equation and get
$$x={e}^{3}$$
.
2. Given the equation
$${2}^{x}=10$$
, solve for
$$x$$
.
First, we use the common logarithm rule to yield
$$x=\log_{2}{10}$$
. Then, we can use what is called the “change of base rule”, which states that
$$\log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}$$
. This gives
$$x=\frac{\log{10}}{\log{2}}$$
. The Rule of
$$10$$
then allows us to convert
$$\log_{10}{}$$
into
$$1$$
to give the final answer of
$$x=\frac{1}{\log{2}}$$
.

# What's Next

Getting comfortable with exponential and logarithmic functions takes patience and practice. Start with our exponential and logarithmic equation practice problems at the top of this page — see how long it takes you to solve them. Need more help? Sign up for Cymath Plus today.
Keep in mind that Cymath can help you tackle everything from Pre-Algebra to Calculus. Make sure you also give other practice problems a try, and best of luck in your studies!