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1st BTL CCSS · Equations and systems Notes 4 Nov 2025

4. Exponential equations

Four types of equations with the unknown in the exponent: $a^x=b$, products and quotients, sums and differences with a common factor, and the "hidden" second-degree ones (change of variable). Four solved exercises with the solution hidden.

What is an exponential equation?

Definition

An exponential equation is an equation in which the unknown appears in the exponent of a power. For example, $\;3^{x} = 81\;$ or $\;5^{x+2} - 5^{x} = 120$.

Warning: not all of them can be solved

Not every exponential equation has a solution that can be found analytically. For example, $\;5^{x} + 2x = 10\;$ cannot be solved for $x$ with algebraic methods — a numerical method would be needed.

Here we will look at the four types of exponential equations that we can solve.

Type 1 $a^{x} = b$

Strategy

We have two options:

  • Factor: if $b$ is a power of $a$ (or can be written as one), we set the exponents equal.
  • Logarithms: otherwise, we take logarithms in base $a$ on both sides.

Example — by factoring

Solve $\;3^{x} = \dfrac{1}{9}$. Since $\dfrac{1}{9} = 3^{-2}$:

$$3^{x} = 3^{-2} \;\Longrightarrow\; x = -2.$$

Example — with logarithms

Solve $\;5^{x} = 7$. Since $7$ is not a power of $5$:

$$\log_{5}(5^{x}) = \log_{5} 7 \;\Longrightarrow\; x = \log_{5} 7 \approx 1.209.$$

Type 2 Products and quotients of exponentials = a number

Strategy

We combine all the exponentials into a single one using the properties:

$$a^{m} \cdot a^{n} = a^{m+n}, \qquad \dfrac{a^{m}}{a^{n}} = a^{m-n}, \qquad (a^{m})^{n} = a^{m\cdot n}.$$

Then we apply the Type 1 methods.

Example

Solve $\;5^{x} \cdot 25^{x+1} : 5^{x-3} = 13$. We convert $25 = 5^{2}$:

$$5^{x} \cdot \bigl(5^{2}\bigr)^{x+1} : 5^{x-3} = 5^{x} \cdot 5^{2x+2} : 5^{x-3} = 13.$$
$$5^{x+(2x+2)-(x-3)} = 5^{2x+5} = 13.$$

We apply $\log_{5}$ to both sides:

$$2x+5 = \log_{5} 13 \;\Longrightarrow\; x = \dfrac{\log_{5} 13 - 5}{2}.$$

Type 3 Sums and differences of exponentials (same base) = a number

Strategy

When all the terms are powers of the same base, we take out the common factor — we use the smallest exponent (whether positive or not) for all the terms — and we are left with a Type 1 equation.

Example

Solve $\;3^{x} - 3^{x-1} + 3^{x+1} = 33$. We take out the common factor $3^{x}$:

$$3^{x}\!\left(1 - \tfrac{1}{3} + 3\right) = 33 \;\Longrightarrow\; 3^{x} \cdot \dfrac{3-1+9}{3} = 33.$$
$$3^{x} \cdot \dfrac{11}{3} = 33 \;\Longrightarrow\; 3^{x} = 9 = 3^{2} \;\Longrightarrow\; x = 2.$$

Type 4 Hidden second-degree equation

Strategy

When $a^{2x}$ and $a^{x}$ (or equivalents) appear in the equation, we make the change of variable $a^{x} = t$. The equation turns into a second-degree one, which we already know how to solve. Then we undo the substitution.

Careful: if we find $t \le 0$, we discard it — because $a^{x}$ is always positive (for $a>0$).

Example

Solve $\;25^{x} - 5^{x+1} + 6 = 0$. We rewrite using $25^{x} = (5^{x})^{2}$ and $5^{x+1} = 5\cdot 5^{x}$:

$$(5^{x})^{2} - 5\cdot 5^{x} + 6 = 0.$$

Substitution $5^{x} = t$:

$$t^{2} - 5t + 6 = 0 \;\Longrightarrow\; t = \dfrac{5 \pm \sqrt{25-24}}{2} = \dfrac{5 \pm 1}{2}.$$

$t_{1} = 3$ and $t_{2} = 2$, both positive. We undo the substitution:

  • If $t = 3$: $\;5^{x} = 3 \Longrightarrow x = \log_{5} 3$.
  • If $t = 2$: $\;5^{x} = 2 \Longrightarrow x = \log_{5} 2$.
$$\boxed{\;x = \log_{5} 2 \quad\text{or}\quad x = \log_{5} 3.\;}$$

Exercises

a Type 1 equation
Solve $\;5^{3x+1} = 1$.

We rewrite $1$ as a power of $5$:  $1 = 5^{0}$. So:

$$5^{3x+1} = 5^{0} \;\Longrightarrow\; 3x+1 = 0.$$
$$\boxed{\;x = -\tfrac{1}{3}.\;}$$

Equivalently, taking $\log_{5}$ on both sides: $\log_{5} 5^{3x+1} = \log_{5} 1 = 0$, which gives the same equation $3x+1=0$.

b Type 2 equation
Solve $\;5^{x+3} \cdot 5^{x} \cdot 5^{x-1} \cdot 5^{-x} = 25$.

We combine all the exponentials by adding the exponents:

$$5^{(x+3)+x+(x-1)+(-x)} = 5^{2}.$$

The exponent simplifies to $\;x+3+x+x-1-x = 2x+2$. Therefore:

$$5^{2x+2} = 5^{2} \;\Longrightarrow\; 2x+2 = 2 \;\Longrightarrow\; \boxed{\,x = 0.\,}$$
c Type 3 equation (common factor)
Solve $\;5^{x+3} + 5^{x} + 5^{x-1} = 131$.

We take out the common factor $5^{x}$:

$$5^{x}\!\left(5^{3} + 1 + \tfrac{1}{5}\right) = 131.$$
$$5^{x}\!\left(125 + 1 + \tfrac{1}{5}\right) = 5^{x} \cdot \dfrac{625 + 5 + 1}{5} = 5^{x} \cdot \dfrac{631}{5} = 131.$$
$$5^{x} = \dfrac{131 \cdot 5}{631} = \dfrac{655}{631}.$$

We apply $\log_{5}$:

$$\boxed{\;x = \log_{5}\!\dfrac{655}{631} \approx 0.0237.\;}$$

A slightly "ugly" answer — when the coefficients don't cancel and $b$ is not a power of $a$, the solution stays in logarithmic form.

d Type 4 equation (change of variable)
Solve $\;25^{x} + 5^{x+1} = 14$.

We rewrite: $\,25^{x} = (5^{x})^{2}\,$ and $\,5^{x+1} = 5 \cdot 5^{x}\,$. So:

$$(5^{x})^{2} + 5 \cdot 5^{x} - 14 = 0.$$

Substitution $5^{x} = t$:

$$t^{2} + 5t - 14 = 0 \;\Longrightarrow\; t = \dfrac{-5 \pm \sqrt{25+56}}{2} = \dfrac{-5 \pm 9}{2}.$$

$t_{1} = 2$  and  $t_{2} = -7$. But $5^{x}$ is always positive, so we discard $t = -7$:

$$5^{x} = 2 \;\Longrightarrow\; \boxed{\;x = \log_{5} 2 \approx 0.431.\;}$$

This is the typical Type 4 trap: the "substitution" leads to a second-degree equation with two solutions, but only the positive ones can be undone.