The equation $a^x=y$ is an example of a general exponent equation (indicial equation) and $2^x = 32$ is an example of a more specific exponential equation (indicial equation).

To solve one of these equations it is necessary to write both sides of the equation with the same base if the unknown is an exponent (index) or with the same exponent (index) if the unknown is the base.

We can solve equations of the form: $x^{\frac{2}{3}} = 4$ as follows.

$$ \begin{align} \displaystyle

(x^{\frac{2}{3}})^{\frac{3}{2}} &= 4^{\frac{3}{2}} &\text{Take the reciprocal power to both sides} \\

x &= (2^2)^{\frac{3}{2}} &\text{The left-hand side becomes } x \\

x &= 2^3 \\

\therefore x &= 8 \\

\end{align} $$

However, when the unknown (or variable) is not a base number but is an index number, a different approach is required.

To attempt to solve exponent equations exactly, express both sides of the equation to the same base and equate the powers.

An exponential equation is an equation in which the unknown occurs as part of the exponent or index.

For example, $2^x = 16$ and $25 \times 3^x = 9$ are both exponential equations.

There are a number of methods we can use to solve exponential equations. These include graphing, using technology, and by using logarithms.

However, in some cases, we can solve algebraically by the following observation:

If $2^x = 8$ then $2^x = 2^3$. Thus $x=3$, and this is the only solution.

If the base numbers are the same, we can equate exponents:

$$a^x = a^n \text{ then } x=n$$

### Example 1

Solve $2^x = 32$.

\( \begin{align} \displaystyle

2^x &= 2^5 \\

\therefore x &= 5 \\

\end{align} \)

Remember to always make the right-hand side equal to zero when solving quadratic equations.

It is a good idea to substitute your answer back into the original equation to check the accuracy of your work.

### Example 2

Solve $3^{x+2} = \dfrac{1}{27}$.

\( \begin{align} \displaystyle

3^{x+2} &= 3^{-3} \\

x+2 &= -3 \\

\therefore x &= -5 \\

\end{align} \)

### Example 3

Solve $4^{x+1} = 32$.

\( \begin{align} \displaystyle

(2^2)^{x+1} &= 2^5 \\

2^{2x+2} &= 2^5 \\

2x+2 &= 5 \\

2x &= 3 \\

\therefore x &= \dfrac{3}{2} \\

\end{align} \)

In some cases, exponential equations (indicial equations) can be expressed in a quadratic form and solved using the Null Factor Law. Look for numbers in exponent form in (index form) similar to $a^{2x}$ or $a^x$ appearing in different terms.

### Example 4

Solve $8^x \times 16^{x+1} = 32$.

\( \begin{align} \displaystyle

(2^3)^x \times (2^4)^{x+1} &= 2^5 \\

2^{3x} \times 2^{4x+4} &= 2^5 \\

2^{3x+4x+4} &= 2^5 \\

2^{7x+4} &= 2^5 \\

7x+4 &= 5 \\

7x &= 1 \\

\therefore x &= \dfrac{1}{7} \\

\end{align} \)

### Example 5

Solve $4^x + 2^x -20 = 0$.

\( \begin{align} \displaystyle

(2^2)^x + 2^x -20 &= 0 \\

(2^x)^2 + 2^x -20 &= 0 \\

(2^x – 4)(x^2 + 5) &= 0 \\

2^x &= 4 \text{ or } 2^x = -5 \\

2^x &= 2^2 &\ 2^x \text{ cannot be negative} \\

\therefore x &= 2 \\

\end{align} \)

Note that, the possible solutions $2^x = -5$ was rejected because there is no value of $x$ for which it will be satisfied. Recall that exponential functions such as $2^x$ are always positive.

Sometimes it is possible to use the methods for solving quadratic equations to help solve exponential equations (indicial equations). Remember $2^{2x} = (2^2)^x$.

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