Exponential Equations

Exponential equations are equations in which the unknown variable appears in the exponent of at least one expression, such as $$ a^x = b $$.

There is no single procedure that solves every exponential equation.

However, when both sides of the equation can be expressed as powers of the same base, we can determine the solution by equating the exponents.

Note. If an exponential equation has exactly one solution it is called a determinate exponential equation. If it has no solutions or infinitely many solutions it is called an indeterminate exponential equation.

A practical example

Consider the exponential equation

$$ 9^{2x} = 27 $$

We rewrite each side using the same base

$$ (3^2)^{2x} = 3^3 $$

$$ 3^{2 \cdot 2x} = 3^3 $$

$$ 3^{4x} = 3^3 $$

Once the bases match, we can directly set the exponents equal to each other.

$$ 4x = 3 $$

Solving for the variable gives

$$ x = \frac{3}{4} $$

This is the solution of the exponential equation.

Since there is exactly one solution, the equation is determinate.

Indeterminate exponential equations

An exponential equation is called indeterminate if it has no solutions or infinitely many solutions.

An exponential equation is solvable only when the exponential expression itself is well defined.

For example, an expression with a positive base raised to any real exponent can never produce a negative value.

$$ 3^x = - 9 $$

This is an instance of an indeterminate exponential equation.

An exponential equation is also indeterminate if it admits infinitely many solutions.

For example, one raised to any real exponent always equals one.

$$ 1^x = 1 $$

This equation holds for every real value of x.

This too is an indeterminate exponential equation.

How to solve exponential equations using logarithms

In many situations exponential equations can be solved efficiently by applying logarithms.

This technique requires both sides of the equation to be positive, since the logarithm of zero or a negative number is undefined.

Example

Consider again the equation

$$ 9^{2x} = 27 $$

Both sides are positive, so the logarithmic method applies.

Take the logarithm base 3 of both sides

$$ \log_3 9^{2x} = \log_3 27 $$

Using the power rule, the exponent can be brought in front

$$ 2x \log_3 9 = \log_3 27 $$

Now isolate x

$$ 2x = \frac{ \log_3 27 }{ \log_3 9 } $$

Since the logarithm base 3 of 27 equals 3

$$ 2x = \frac{ 3 }{ \log_3 9 } $$

and the logarithm base 3 of 9 equals 2

$$ 2x = \frac{ 3 }{ 2 } $$

Finally, divide both sides by two

$$ 2x \cdot \frac{1}{2} = \frac{ 3 }{ 2 } \cdot \frac{1}{2} $$

$$ x = \frac{ 3 }{ 4 } $$

The result matches the previous method.

And so on.