Exponential equations by substitution: ranges and back-substitution
Choose a substitution from the bases, restrict its range, retain denominator exclusions, and recover every original-variable solution.
Content updated: October 22, 2026
A suitable substitution collects several powers into an algebraic equation. Filter its algebraic roots using the substitution's range, then recover the original variable and check every condition of the original equation.
Objectives: choose a substitution from relations between bases; check positivity, nonzero denominators, and the range for back-substitution; solve an example with a squared exponent.
Prerequisites: quadratic equations, real powers, logarithms, and the range of an exponential function.
On this page: Idea · Method · Examples · Practice · Further learning
1. The range of a substitution
For $a>0$, $a\ne1$, and $x$ ranging over $\mathbb R$, the substitution $t=a^x$ satisfies $t>0$, and every positive value is attained exactly once. Thus discard $t\le0$; each root $t_0>0$ gives $x=\log_a t_0$.
For $t=a^{u(x)}$, the condition $t>0$ may be insufficient. For example, if $a>1$ and $u(x)=(x-1)^2\ge0$, require $t\ge1$. On back-substitution, solve $(x-1)^2=\log_a t_0$ and check that the right side is nonnegative.
2. Recognizing the structure before substituting
The pair $a^{2x},a^x$ suggests $t=a^x$. For $4^x,6^x,9^x$, divide by $9^x>0$ to obtain $(2/3)^{2x},(2/3)^x$. For reciprocal conjugate bases, first use a common base and then substitute.
If the equation has a denominator, retain its condition after substitution; multiplying through does not reinstate a value that makes the denominator 0. If both $x$ and $t$ remain, only the notation has changed: the equation is not automatically in one unknown. Check the solutions against both the substitution and the original equation.
3. Four worked examples
Example 1. A polynomial in 2ˣ
Solve $4^x-4\cdot2^x+3=0$. Set $t=2^x>0$ to obtain $(t-1)(t-3)=0$. Both roots 1 and 3 are valid. Hence $x=0$ or $x=\log_2 3$.
Example 2. Retaining a denominator condition
Solve $(8^x+2^x)/(4^x-2^x)=5$. The denominator is $2^x(2^x-1)$, so require $x\ne0$. Set $t=2^x>0$, $t\ne1$. Divide numerator and denominator by $2^x>0$; the equation becomes
Accept $t=2,3$, since both differ from 1. Thus $x=1$ or $x=\log_2 3$, both different from 0. Substituting either $t=2$ or 3 into the fraction gives 5.
Example 3. Three bases related by squares
Solve $3\cdot4^x-5\cdot6^x+2\cdot9^x=0$. Divide by $9^x>0$ and set $t=(2/3)^x>0$ to obtain $3t^2-5t+2=(3t-2)(t-1)=0$. Accept $t=1$ or $t=2/3$, giving $x=0$ and $x=1$, respectively.
Example 4. A squared exponent
Solve $(2+\sqrt3)^{(x-1)^2}+(2-\sqrt3)^{x^2-2x-1}=4/(2-\sqrt3)$. Set $B=2+\sqrt3>1$. Since $2-\sqrt3=1/B$ and $x^2-2x-1=(x-1)^2-2$, the substitution $t=B^{(x-1)^2}\ge1$ transforms the equation into
We have $B^2=7+4\sqrt3$ and $B^2+1=4B$, so the polynomial factors as $(t-1)(t-B^2)$. Both roots 1 and $B^2$ satisfy $t\ge1$. For $t=1$, injectivity implies $(x-1)^2=0$, so $x=1$. For $t=B^2$, $(x-1)^2=2$, so $x=1\pm\sqrt2$. Thus the solutions are $1-\sqrt2,1,1+\sqrt2$.
4. Practice exercises
- Solve $9^x-4\cdot3^x-5=0$.
- Solve $2^x+2^{-x}=2$.
- Solve $2^{x^2}=1/2$.
Show hints and answers
- Set $t=3^x>0$: $(t-5)(t+1)=0$. Discard $-1$ and accept 5, so $x=\log_3 5$.
- Set $t=2^x>0$ to obtain $t+1/t=2$, equivalent to $(t-1)^2=0$. Hence $x=0$.
- Since $x^2\ge0$, the left side is at least 1, so there is no solution. A substitution must retain $t=2^{x^2}\ge1$ to exclude 1/2.
5. Common mistakes
Do not state only t>0 when the actual range is smaller. Reject substituted roots violating denominator conditions. After returning to x, solve the full exponent equation; a single t value can correspond to several x values.
6. Related lessons
- Powers (English lesson)
- Exponential functions (English lesson)
- Exponential equations using a common base (English lesson)
- Exponential equations by taking logarithms (English lesson)
- Exponential equations using function properties (English lesson)
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