Solving exponential equations

A schematic showing the change in population of rabbits from February to May. In February, there are two rabbits. In March, there are four. In April, there are 8. By May, there are 16. This is represented by the equation N equals 2 times 2 to the power of t. t represents time. — Rabbits reproduce so quickly that they can be considered to breed at exponential rates.

Exponential equations are common in fields such as science, finance, and population dynamics, where exponential growth or decay is observed. By learning how to isolate variables and find solutions to exponential equations, you will be able to handle a variety of real-world problems involving exponential relationships.

Exponential equations

Indicial (or exponential) equations have a variable in the power or exponent. The general form of an exponential equation is: $a^{x} = b$ . Examples include:

$5^{x} = 17$
$2^{1 - x} = 4^{3 x}$
$A = P (1 + \frac{r}{n})^{n t}$

If we can write $b$ as a number with a base $a$ and an index $x$ , then we can use a set of rules to find $x$ .

If two equal numbers have a common base $(a)$ , then the indices $(x)$ must also be equal.
The logarithm of both sides can be taken to solve for $x$ .

Mathematically, these rules can be represented as:

$\begin{aligned} a^{x} = a^{y} & \Rightarrow x = y (1) \\ a^{x} = b \Rightarrow & \log_{y} a^{x} = \log_{y} b (2) \end{aligned}$

The rule you use to solve the exponential equation depends on whether it has integer or non-integer solutions.

Integer solutions

The first rule comes in handy for exponential equations that have integer solutions. You can quickly tell whether to use rule $1$ by checking whether you can write the number on the other side of the equal sign, i.e. the $b$ component, as an exponential with the same base. Let's look at some examples.

Example 1 – solving exponential equations with integer solutions

Solve $3^{x} = 27$ .

We should first check whether $27$ can be written to the same base as the exponential expression. In this case, the base is $3$ , so we need to check if $27$ can be written as an exponential with the same base. It can be written as $27 = 3^{3}$ . Thus:

$\begin{aligned} 3^{x} & = 27 \\ 3^{x} & = 3^{3} \end{aligned}$

These expressions have the same base, so their indices must be equal.

$x = 3$

Solve $2^{1 - x} = \frac{1}{8}$ .

We should first check whether $\frac{1}{8}$ can be written to the same base as the exponential expression. In this case, the base is $2$ , so we need to check if $\frac{1}{8}$ can be written as an exponential with the same base. It can be written as $\frac{1}{8} = 2^{- 3}$ . Thus:

$\begin{aligned} 2^{1 - x} & = \frac{1}{8} \\ 2^{1 - x} & = 2^{- 3} \end{aligned}$

These expressions have the same base, so their indices must be equal.

$\begin{aligned} 1 - x & = - 3 \\ x & = 1 + 3 \\ x & = 4 \end{aligned}$

Solve $5^{\frac{1}{x}} = 25$ .

We should first check whether $25$ can be written to the same base as the exponential expression. In this case, the base is $5$ , so we need to check if $25$ can be written as an exponential with the same base. It can be written as $25 = 5^{2}$ . Thus:

$\begin{aligned} 5^{\frac{1}{x}} & = 25 \\ 5^{\frac{1}{x}} & = 5^{2} \end{aligned}$

These expressions have the same base, so their indices must be equal.

$\begin{aligned} \frac{1}{x} & = 2 \\ x & = \frac{1}{2} \end{aligned}$

Non-integer solutions

Not all equations have integer solutions. That is, the solutions may not be whole numbers. For example: $3^{x} = 10$ has a solution between $2$ and $3$ , since $3^{2} = 9$ and $3^{3} = 27$ .

Here, we must use rule $2$ to solve the equation.

Remember that for $a^{x} = b$ :

$\log_{y} a^{x} = \log_{y} b \Rightarrow x \log_{y} a = \log_{y} b$

These types of equations can be solved using a calculator that can do logarithms with base $10$ or base $e$ . This means that $y = 10$ or $y = e$ .

On your calculator, the LOG button will calculate $\log_{10} b$ and the LN button will calculate $\log_{e} b$ . Logarithms with base $e$ are known as natural logarithms and sometimes the abbreviation $\ln b$ is used for $\log_{e} b$ .

Example 1 – solving exponential equations with non-integer solutions

Solve $3^{x} = 10$ to three decimal places.

\begin{aligned} 3^{x} & = 10 take \log_{10} of both sides \\ \log_{10} 3^{x} & = \log_{10} 10 rearrange using the third logarithm law \\ x \log_{10} 3 & = 1 \\ x & = \frac{1}{\log_{10} 3} \\ = 2.096 \end{aligned}

Solve $2 \times 5^{x + 1}$ to two decimal places.

\begin{aligned} 2 \times 5^{x + 1} & = 15 \\ 5^{x + 1} & = \frac{15}{2} \\ 5^{x + 1} & = 7.5 take \log_{10} of both sides \\ \log_{10} 5^{x + 1} & = \log_{10} 7.5 rearrange using the third logarithm law \\ (x + 1) \log_{10} 5 & = \log_{10} 7.5 \\ x + 1 & = \frac{\log_{10} 7.5}{\log_{10} 5} \\ x + 1 & = 1.25 \\ x & = 0.25 \end{aligned}

Solve $2^{2 x + 1} = 5^{2 - x}$ to three decimal places.

\begin{aligned} 2^{2 x + 1} & = 5^{2 - x} take \log_{10} of both sides \\ \log_{10} 2^{2 x + 1} & = \log_{10} 5^{2 - x} rearrange using the third logarithm law \\ (2 x + 1) \log_{10} 2 & = (2 - x) \log_{10} 5 \\ (2 x + 1) 0.301 & = (2 - x) 0.699 \\ 0.602 x + 0.301 & = 1.398 - 0.699 x \\ 1.301 x & = 1.097 \\ x & = 0.843 \end{aligned}

Solve $\log_{10} (2 x + 1) = \log_{10} 3$ .

\begin{aligned} \log_{10} (2 x + 1) & = \log_{10} 3 \\ 2 x + 1 & = 3 \\ x & = 1 \end{aligned}

Solve $2^{x - 3} = 10$ .

\begin{aligned} 2^{x - 3} & = 10 take log of both sides \\ \log_{10} (2^{x - 3}) & = \log_{10} 10 \\ (x - 3) \log_{10} 2 & = 1 \\ (x - 3) (0.301) & = 1 \\ x - 3 & = \frac{1}{0.301} \\ x & = \frac{1}{0.301} + 3 \\ = 6.32 \end{aligned}

Exercise – solving exponential equations

Solve for $x$ .

$3^{1 - x} = 27$
$2^{2 x - 1} = 128$
$9^{\frac{1}{x}} = 3^{- 4}$
$5^{x} = 12$
$2^{x - 3} = 9$
$\frac{1}{2^{x + 1}} = 4^{x + 2}$
$\log_{10} x = \log_{10} 4 - \log_{10} 2$
$\log_{2} 2 x - 2 \log_{2} 3 = \log_{2} 6$
$10^{x + 1} = 7$

$x = - 2$
$x = 4$
$x = - \frac{1}{2}$
$x = 1.54$
$x = 6.17$
$x = - 1.67$
$x = 2$
$x = 27$
$x = - 0.155$

Growth and decay

Growth and decay are often modelled using exponential and logarithmic expressions, across all branches of STEM.

Example – solving problems involving growth and decay

The number of bacteria present in a sample is given by $N = 800 e^{0.2 t}$ , where $t$ is the time in seconds. Find:

the initial number of bacteria

the time taken for the bacteria count to reach $10000$ .

For part a, substitute $t = 0$ into the equation.

$\begin{aligned} N & = 800 e^{0.2 t} \\ = 800 e^{0.2 \times 0} \\ = 800 e^{0} \\ = 800 \end{aligned}$

The initial bacteria count is 800.

For part b, substitute $N = 10000$ into the equation.

\begin{aligned} N & = 800 e^{0.2 t} \\ 10000 & = 800 e^{0.2 t} \\ \frac{10000}{800} & = e^{0.2 t} \\ 12.5 & = e^{0.2 t} take \ln of both sides \\ \ln 12.5 & = \ln e^{0.2 t} rearrange using the third logarithm law \\ \ln 12.5 & = 0.2 t \ln e recall that \ln e = \log_{e} e = 1 \\ \ln 12.5 & = 0.2 t \\ t & = \frac{\ln 12.5}{0.2} \\ = 12.6 \end{aligned}

It takes $12.6$ sec for the number of bacteria to reach $10000$ .

Exercise – solving problems involving growth and decay

The decay rate $R$ for a radioactive element is given by $R = 400 e^{- 0.03 t}$ , where $t$ is measured in seconds. Find:

the initial decay rate
the time for the decay rate to reduce to half of the initial decay rate.

The charge units $Q$ on the plate of a condenser $t$ seconds after it starts to discharge is given by $Q = Q_{0} 10^{- k t}$ , where $Q_{0}$ is the initial charge, $t$ is the time in seconds, and $k$ is a constant.
If the intial charge is $5076$ units and $Q = 1840$ when $t = 0.5$ seconds, find:

the value of $k$
the time needed for the charge to fall to $1000$ units
the charge after $2$ sec.

1. $400$ units per second
2. $23.1$ seconds
1. $0.881$
2. $0.8$ seconds
3. $87.6$ charge units

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Solving exponential equations