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Solving exponential and logarithmic equations

Index laws and the laws of logarithms are essential tools for simplifying and manipulating exponential and logarithmic functions.

There is an inverse relationship between exponential and logarithmic functions. That is, each function effectively 'undoes' what the other does. This fact provides insight into how these functions interact and complement each other. This inverse relationship is especially useful when solving exponential and logarithmic equations.

Use this page to revise the following concepts of exponential and logarithmic equations:

Index laws
Laws of logarithms
Solving exponential equations
Solving logarithmic equations

Index laws

In mathematics, an index (plural indices) is the power or exponent to which a base is raised. It can be either a number or a variable. For example, in the number \( 2^3 \), the index is 3 and the base is 2. The exponent 3 tells us to multiply 2 by itself three times.

\[2^3=2\times2\times2=8\]

The Index Laws are rules that describe how exponents behave when we multiply and divide terms in index form, or when we raise one power to another. They provide a systematic way to simplify expressions and solve equations involving exponents. These laws are applied when performing algebraic operations involving indices and solving algebraic equations.

When multiplying powers with the same base, add the indices.
\[ a^m \times a^n = a^{m+n} \]
For example,
\[ a^2 \times a^9 = a^{2+9} = a^{11} \]
When dividing powers with the same base, subtract the indices.
\[ \frac{a^m}{a^n} = a^{m-n}, \text{ where } a \neq 0 \]
For example,
\[ \frac{a^{11}}{ a^3} = a^{11-3} = a^8 \]
When raising a power to an index, multiply the indices.
\[ (a^m)^n = a^{m \times n} \]
For example,
\[ (a^2)^5 = a^{2 \times 5} = a^{10} \]
Any non-zero value raised to the power of 0 is equal to 1.
\[ a^0 = 1,\text{ where } a \neq 0 \]
For example,
\[ 4^0 = 1\]
When an index (exponent) is negative, place the expression under 1 and rewrite the exponent as a positive value. In other words, take the reciprocal of the expression to remove the negative exponent.
\[
a^{-m} = \frac{1}{a^m},\text{ where } a \neq 0
\]
For example,
\[ a^{-7} = \frac{1}{a^7} \]
When the exponent is a fraction, first take the root indicated by the denominator, then raise the result to the power of the numerator.
\[ a^{\frac{n}{m}} = \left( \sqrt[m]{a} \right)^n \text{ or } a^{\frac{n}{m}} = \sqrt[m]{a^n} \]
For example,
\[ a^{\frac{3}{5}} = \left( \sqrt[5]{a} \right)^3 \text{ or } a^{\frac{3}{5}} = \sqrt[5]{a^3} \]

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Laws of logarithms

Logarithms are the inverse of exponents: that is, if \( a^n = m \), then \( \log_a(m) = n \).

The equation \( \log_a(m) = n \), means that the base \(a\), when raised to the power of \(n\), equals the argument \(m\). Alternatively, it is asking how many times \(a\) must be multiplied by itself to equal \(m\). The argument \(m\) must be positive, and the base \(a\) must be positive and cannot equal 1. Thus, the domain of \( \log_a(m) \),is \( m > 0 \).

Understanding index laws is crucial for converting from exponential to logarithmic forms. For example, since \( 10^3 = 1000 \), it follows that \( \log_{10} 1000 = 3 \). Like the laws of indices, the laws of logarithms are used to simplify and rearrange more complicated logarithmic expressions.

The logarithm of the product of two arguments is the sum of logarithms of each argument.
\[ \log_a(m \times n) = \log_a(m) + \log_a(n) \text{ where } a, m, n \in \mathbb{R}^+ \text{ and } a \neq 1 \]
This law can also be written in reverse.
\[ \log_a(m) + \log_a(n) = \log_a(m \times n) \text{ where } a, m, n \in \mathbb{R}^+ \text{ and } a \neq 1 \]
For example,
\[ \log_2(3 \times 5) = \log_2(3) + \log_2(5) \text{ or } \log_2(3) + \log_2(5) = \log_2(3 \times 5) \]
The logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator.
\[ \log_a\left(\frac{m}{n}\right) = \log_a(m) - \log_a(n) \text{ where } a, m, n \in \mathbb{R}^+ \text{ and } a \neq 1 \]
This law can also be written in reverse.
\[ \log_a(m) - \log_a(n) = \log_a\left(\frac{m}{n}\right) \text{ where } a, m, n \in \mathbb{R}^+ \text{ and } a \neq 1 \]
For example,
a. \[ \log_5\left(\frac{21}{3}\right) = \log_5(21) - \log_5(3) \]
b. \([\log_5(100) - \log_5(4) = \log_5\left(\frac{100}{4}\right) = \log_5(25) = 2 \]
The logarithm of an exponential number is the exponential times the logarithm of the base.
\[
\log_a(m^n) = n \times \log_a(m) \text{ where } a, m \in \mathbb{R}^+ \text{ and } a \neq 1
\]
This law could also be written in reverse.
\[
n \log_a(m) = \log_a(m^n) \text{ where } a, m \in \mathbb{R}^+ \text{ and } a \neq 1
\]
For example,
a. \( \log_2(4^3) = 3 \log_2(4) \)
b. \( 3 \log_2(10) = \log_2(10^3) = \log_2(1000) \)
The logarithm of 1 to any base is always equal to zero, as long as \(a\) is positive and \( a \neq 1 \).
\[
\log_a(1) = 0 \text{ where } a \in \mathbb{R}^+ \text{ and } a \neq 1
\]
For example,
\[
\log_4(1) = 0
\]
The logarithm of the argument (inside the parenthesis) wherein the argument equals the base is equal to 1 for \( 1 \text{ for } a \in \mathbb{R}^+ \text{ and } a \neq 1 \).
\[
\log_a(a) = 1 \text{ where } a \in \mathbb{R}^+ \text{ and } a \neq 1
\]
For example,
\[
\log_5(5) = 1
\]
When the reciprocal of the input to a logarithm is taken, the result is equal to -1 multiplied by the logarithm of the original input.
\[
\log_a\left(\frac{1}{m}\right) = \log_a(m^{-1}) = -\log_a(m) \text{ where } a \in \mathbb{R}^+, a \neq 1 \text{ and } m \neq 0
\]
For example,
\[
\log_2\left(\frac{1}{3}\right) = \log_2(3^{-1}) = -\log_2(3)
\]
The logarithm of an exponential number where its base is the same as the base of the logarithm is equal to the component.
\[
\log_a(a^m) = m \text{ where } a \in \mathbb{R}^+ \text{ and } a \neq 1
\]
For example,
\[
\log_2(2^3) = 3
\]
To convert a logarithmic expression from one base to a log-fraction expression with a different base.
\[
\log_a(m) = \frac{\log_b(m)}{\log_b(a)} \text{ where } a, b, m \in \mathbb{R}^+ \text{ and } a \neq 1
\]
For example,
\[
\log_2(5) = \frac{\log_e(5)}{\log_e(2)}
\]

Note

A logarithm can have any positive value other than 1 as its base, but two particular bases are generally regarded as being especially important:

Log with a base of 10
Log with a base of Euler’s number \(e\) (approximately 2.71828).

A base of 10 is sometimes referred to as the common logarithm.

Logs with a base of number \(e\) are called the Natural Logarithm. The natural logarithm is sometimes written as \( \ln(x) \).

Occasionally, logarithms are written without a base, for example \( \log(3) \). Depending on the context, this can refer to the common logarithm \(\left(\log_{10}(3)\right)\) or the natural logarithm \(\left(\ln(3)\right)\).

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Solving exponential equations

Exponential equations, as the name suggests, involve exponents. The exponent of a number (also known as base) indicates the number of times the base is multiplied. However, the power can also be a variable instead of a number. When it appears as part of an equation, it is called an exponential equation.

There are three types of exponential equations. They are as follows:

Equations with the same bases on both sides, for example \( 2^x = 2^3 \)
Equations with different bases that can be made the same, for example: \( 3^x = 27 \) which can be written as \( 3^x = 3^x \)
Equations with different bases that cannot be made the same, for example: \( 4^x = 3 \).

Equations with the same bases on both sides
Equations with different bases that can be made the same
Equations with different bases that cannot be made the same

While solving an exponential equation, the bases on both sides might be the same. If the bases on both sides are the same, the one-to-one property will be used to solve for unknown variables.

One-to-one property

If \( a^m = a^n \),

then \( m = n \)

where \( a \in \mathbb{R}^+ \text{ and } a \neq 1 \)

To solve for unknown variables if the exponential equation is in the form of \( a^m = a^n \) where \( a \in \mathbb{R}^+ \text{ and } a \neq 1 \):

Use the one-to-one property to equate the exponents
Solve the resulting equation for the unknown variables

Worked Example 1

Solve the following equation.

\[ 5^x = 5^8 \]

\[ x=8\]

Worked Example 2

Solve the following equation.

\[
2^{2x-5} = \frac{2^{3x}}{2}
\]

Rewrite \(2\) as \(2^1\)

\[
2^{2x-5} = \frac{2^{3x}}{2^1}
\]

Use the division law of indices

\[
2^{2x-5} = 2^{3x-1}
\]

Apply the one-to-one property

\[
2x - 5 = 3x - 1
\]

Subtract \(3x\) and add \(5\) to both sides

\[ -x = 4 \]

Divide both sides by \(-1\)

\[ x = -4 \]

Sometimes the base for an exponential equation is not explicitly shown or the bases on both sides are different, for example, \(2^x = 128\). In these cases, to rewrite the terms in the equation as powers with the same base and solve it using one-to-one property is recommended.

To solve for unknown variables if the bases on both sides are not the same or the base is not explicitly shown:

Rewrite each side in the equation as a power with the same base
Use index laws to simplify, if necessary, so that the resulting equation has the form \(a^m = a^n\).
Use the one-to-one property to equate the exponents.
Solve the resulting equation for the unknown variables.

Worked Example 1

Solve the following equation.

\[ 3^{2x-2} = 81\]

Rewrite 81 as \(3^4\)

\[ 3^{2x-2} = 3^4\]

Apply the one-to-one property

\[2x - 2 = 4\]

Solve for \(x\)

\[ x=3\]

Worked Example 2

Solve the following equation.

\[ 8^{x+2} = 16^{x+1}\]

Write \(8\) and \(16\) as powers of \(2\)

\[ (2^3)^{x+2} = (2^4)^{x+1}\]

Use power of a power law

\[ 2^{(3x+6)} = 2^{(4x+4)}\]

Apply the one-to-one property

\[ 3x + 6 = 4x + 4\]

Solve for \(x\)

\[ x=2\]

When the bases in an exponential equation cannot be made the same, taking the logarithm of each side is often the best way to solve it. For instance, in the equation \(2^x = 3\), there’s no simple way to express both sides with a common base, so logarithms are used instead. The following formula is used:

Bases cannot be made the same

If \( a^x = n \),

then \( \log_a(a^x) = \log_a(n) \)

thus \( x = \log_a(n) \)

where \( a \in \mathbb{R}^+ \text{ and } a \neq 1 \)

To solve for unknown variables if the bases on both sides are different and cannot be made the same.

Apply the logarithm of both sides of the equation
- If one of the terms in the equation has base 10, use the common logarithm (base of 10)
- If none of the terms in the equation has base 10, use the natural logarithm (base of \(e\))
Use the laws of logarithms to solve for the unknown variables

Worked Example 1

Solve the following equation.

\[ 5^x = 4 \]

Take \( \log_5 \) of each side

\[ \log_5(5^x) = \log_5(4) \]

Use the identity law to solve for \(x\)

\[ x = \log_5(4) \]

Worked Example 2

Solve the following equation.

\[ 3^{x+2} = 4^x \]

Take natural log of both sides

\[ \log_e(3^{x+2}) = \log_e(4^x) \]

Use power rule of logarithms

\[ (x+2) \log_e 3 = x \log_e 4 \]

Expand the bracket

\[ x \log_e 3 + 2 \log_e 3 = x \log_e 4 \]

Rearrange the equation

\[ x \log_e 3 - x \log_e 4 = -2 \log_e 3 \]

Use quotient and reciprocal rules of logarithms

\[ x \log_e\left(\frac{3}{4}\right) = \log_e\left(\frac{1}{9}\right) \]

Divide both sides by the coefficient of \(x\)

\[ x = \frac{\log_e\left(\frac{1}{9}\right)}{\log_e\left(\frac{3}{4}\right)} \]

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Solving logarithmic equations

Logarithmic equations are equations which incorporate logarithms.

Generally, there are two types of logarithmic equations.

A single logarithm on each side of the equation having the same base. If \( \log_a(m) = \log_a(n) \), then m = \(n\). For example, \( \log_2(2x) = \log_2(4) \) implies that \(2x = 4\).
A single logarithm on one side of the equation. \( \log_a(m) = b\) can be rewritten as \(a^b = m\). For example, \( \log_2(2x) = 2 \) becomes \(2^2 = 2x\).

A single logarithm on each side of the equation with same bases
A single logarithm on one side of the equation

If the bases on both sides are the same, the one-to-one property can be used to solve for unknown variables.

One-to-one property

If \( \log_a(m) = \log_a(n) \),

then \( m = n \),

where \( a,n,m \in \mathbb{R}^+ \text{ and } a \neq 1 \)

Worked Example 1

Solve the following equation.

\[\begin{aligned}
\log_3(x) &= \log _3 (9) \\ 3x&=9 \\ x&=3
\end{aligned}\]

Worked Example 2

Solve the following equation.

\[\log _2(x+2)+\log _2(3)=\log _2(27)\]

Use Product Rule in reverse to condense the left-hand side

\[\log _2 3(x+2)=\log _2(27)\]

Apply the one-to-one proerty

\[3(x+2)=27\]

Expand the bracket

\[3x+6=27\]

Subtract \(6\) from both sides

\[3x=21\]

Divide both sides by the coefficient \(3\)

\[x=7\]

Worked Example 3

Solve the following equation.

\[
\log_3(x) + \log_3(x - 2) = \log_3(x + 10)
\]

Use Product Rule in reverse to condense the left-hand side

\[ \log_3 x(x - 2) = \log_3(x + 10) \]

Apply the one-to-one property

\[ x(x - 2) = (x + 10) \]

Expand the bracket

\[ x^2 - 2x = x + 10 \]

Move \(x\) and \(10\) to the left-hand side

\[ x^2 - 3x - 10 = 0 \]

Use factoring or cross-method to factorise the left-hand side

\[ (x + 2)(x - 5) = 0 \]

Use Null Factor Law to solve for \(x\)

\[ x = -2 or x = 5 \]

Check the domain of the logarithms

As the domain must be positive, then

\[ x > 2 \]

Find the valid solution

\[ x = 5 \]

Note

The most important step when solving logarithmic equations is to check if all values can be taken as solutions.

The logarithm of a negative number and the logarithm of zero are both undefined; therefore, only values of \(x\) that result in a positive logarithm are valid.

In the example above, if \(x = 2\) the the equation would read \( \log_3(2) + \log_3(2-2) \); however, \( \log_3(0) \) is undefined, so \(x > 2\).

When there is only one logarithm on one side, rewriting the equation in its exponential form is often the simplest way to solve it.

If \(\log _a(m)=n\),

then \(a^n=m\),

where \(a, m \in \mathbb{R}^{+}\) and \(a \neq 1\)

Worked Example 1

Solve the following equation.

\[
\log_3(8x - 15) = 4
\]

Convert to the exponential form

\[ 3^4 = 8x - 15 \]

Evaluate the left-hand side

\[ 81 = 8x - 15 \]

Add \(15\) to both sides

\[ 96 = 8x \]

Divide both sides by \(8\)

\[ x = 12 \]

Worked Example 2

Solve the following equation.

\[
\log_2(x^2 - 4x) = \log_2(4 - x) + 5)
\]

Gather all logarithmic expressions to the left-hand side

\[
\log_2(x^2 - 4x) 0- \log_2(4 - x) = 5)
\]

Use the Quotient Rule to condense the left-hand side

\[ \log_2\left(\frac{x^2 - 4x}{4 - x}\right) = 5 \]

Convert to the exponential form

\[ \frac{x^2 - 4x}{4 - x} = 2^5 \]

Multiple \(4 - x\) to get rid of the fraction

\[ x^2 - 4x = 32(4 - x) \]

Expand the bracket

\[ x^2 - 4x = 128 - 32x \]

Rearrange the equation

\[ x^2 + 28x - 128 = 0 \]

Use Null Factor Law to solve for \(x\)

Check the domain of the logarithms

\[ x < 0 \]

Find the valid solution

\[ x = -32 \]

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Solving exponential and logarithmic equations

Index laws

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Laws of logarithms

Note

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Solving exponential equations

Worked Example 2

Worked Example 1

Worked Example 2

Worked Example 1

Worked Example 2

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Solving logarithmic equations

Worked Example 1

Worked Example 2

Worked Example 3

Note

Worked Example 1

Worked Example 2

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