## Exponential Equations

Exponential equations have the unknown variable in the exponent. Here are some examples:

\begin{align*} {3}^{x + 1} & = 9 \\ {5}^{t} + 3 \times {5}^{t – 1}& = 400 \end{align*}

If we can write a single term with the same base on each side of the equation, we can equate the exponents. This is one method to solve exponential equations.

**Important:** if \(a>0\) and \(a\ne 1\) then:

\begin{align*} {a}^{x}& ={a}^{y} \\ \text{then } x& = y \text{ (same base)} \end{align*}

Also notice that if \(a = 1\), then \(x\) and \(y\) can be different.

## Example

### Question

Solve for \(x\): \({3}^{x + 1}=9\).

### Change the bases to prime numbers

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

### The bases are the same so we can equate exponents

\begin{align*} x+1& = 2 \\ \therefore x & = 1 \end{align*}

## Example

### Question

Solve for \(t\): \({3}^{t}=1\).

### Solve for \(t\)

We know from the exponent identities that \(a^{0}=1\), therefore:

\begin{align*} {3}^{t}& = 1 \\ 3^{t} & = 3^{0} \\ \therefore t &= 0 \end{align*}

## Example

### Question

Solve for \(t\): \({5}^{t} + 3\cdot {5}^{t + 1} = 400\).

### Rewrite the expression

\[{5}^{t} + 3({5}^{t} \cdot 5) = 400\]

### Take out a common factor

\begin{align*} {5}^{t}(1+3\cdot5)&=400 \\ {5}^{t}(1 + 15) &= 400 \end{align*}

### Simplify

\begin{align*} {5}^{t}(16) & = 400 \\ {5}^{t} & =25 \end{align*}

### Change the bases to prime numbers

\[{5}^{t} = {5}^{2}\]

### The bases are the same so we can equate exponents

\[\therefore t=2\]

## Example

### Question

Solve for \(x\): \[3^{2x}-80\cdot 3^{x} – 81 = 0\]

### Factorise the trinomial

\[(3^{x}-81)(3^{x}+1)=0\]

### Solve for \(x\)

\(3^x = 81\) or \(3^x = -1\). However \(3^x = -1\) is undefined, so:

\begin{align*} 3^{x}& = 81 \\ {3}^{x} & = 3^{4} \\ x & = 4 \end{align*}

Therefore \(x = 4\)

## Example

### Question

Solve for \(p\): \[p-13{p}^{\frac{1}{2}} + 36 = 0\]

### Rewrite the equation

We notice that \({({p}^{\frac{1}{2}})}^{2} = p\) so we can rewrite the equation as:

\[{({p}^{\frac{1}{2}})}^{2} – 13{p}^{\frac{1}{2}} + 36 = 0\]

### Factorise as a trinomial

\[({p}^{\frac{1}{2}} – 9)({p}^{\frac{1}{2}} – 4) = 0\]

### Solve to find both roots

\(\begin{array}{rlcrl} {p}^{\frac{1}{2}} – 9& = 0 & \text{ or } & {p}^{\frac{1}{2}} – 4 & =0 \\ {p}^{\frac{1}{2}} & =9 & & {p}^{\frac{1}{2}} & =4 \\ {({p}^{\frac{1}{2}})}^{2} & = {(9)}^{2} & & {({p}^{\frac{1}{2}})}^{2} & = {(4)}^{2} \\ p & =81 & & p & =16 \end{array}\)

Therefore \(p = 81\) or \(p = 16\).

Learners may find Worked Example 13 much easier using the \(k\)-substitution method. You may choose to return to this example once the \(k\)-substitution has been taught.

The solution using \(k\)-substitution is as follows:

\begin{align*} 2^{x}-2^{4-x} &=0 \\ 2^{x}-2^{4} \cdot 2^{-x} & = 0\\ 2^{x}-\dfrac{2^{4}}{2^{x}} & = 0 \\ \text{Let } 2^{x}&=k\\ k-\dfrac{2^{4}}{k} &=0\\ \times k \qquad &k^{2}-16 = 0 \\ (k-4)(k+4)=0\\ k = -4 & \text{ or } \qquad k = 4 \\ 2^{x} \ne -4 & \qquad 2^{x}=4 \\ & \qquad 2^{x}=2^{2}=4 \\ x& = 2 \end{align*}

## Example

### Question

Solve for \(x\): \[2^{x}-2^{4-x} = 0\]

### Rewrite the equation

In order to get the equation into a form which we can factorise, we need to rewrite the equation:

\begin{align*} 2^{x}-2^{4-x} &= 0 \\ 2^{x}-2^{4} \cdot 2^{-x} &= 0 \\ 2^{x}-\cfrac{2^{4}}{2^{x}} &= 0 \end{align*}

Now eliminate the fraction by multiplying both sides of the equation by the denominator, \(2^{x}\).

\begin{align*} (2^{x}-\cfrac{2^{4}}{2^{x}}) \times 2^{x}&= 0 \times 2^{x} \\ 2^{2x}-16 &= 0 \end{align*}

### Factorise the equation

Now that we have rearranged the equation, we can see that we are left with a difference of two squares. Therefore:

\begin{align*} 2^{2x}-16 &= 0 \\ (2^x – 4)(2^x+4) &=0 \\ 2^x & = 4 \qquad 2^x \ne -4 \quad \text{(a positive integer with an exponent is always positive)}\\ 2^{x}=2^{2} & =4 \\ x& = 2 \end{align*}

Therefore \(x = 2\).