Some exponential equations can be solved without logarithms: if both sides can be written as powers of the same base, the exponents must match. You will use the exponent rules and the shape of the exponential graph.
Idea clave
Rewrite both sides as powers of one base $b$ (with $b > 0$, $b \ne 1$). Then $b^m = b^n$ means $m = n$. Set the exponents equal and solve.
Por qué funciona
An exponential function $b^x$ with $b > 0$, $b \ne 1$ is always increasing or always decreasing, so it never takes the same value twice. It is one-to-one, which means equal outputs come from equal inputs. That is exactly the step $b^m = b^n \Rightarrow m = n$.
Worked examples
Ejemplo resuelto: A direct match
Solve $2^{x + 1} = 32$.
- $32 = 2^5$, so $2^{x + 1} = 2^5$.
- $x + 1 = 5$.
Respuesta: $x = 4$
Ejemplo resuelto: Different bases, same family
Solve $4^{x} = 8^{x - 1}$.
- Use base 2: $4 = 2^2$ and $8 = 2^3$, so $2^{2x} = 2^{3(x - 1)}$.
- $2x = 3x - 3$.
Respuesta: $x = 3$
Ejemplo resuelto: A fraction and a negative exponent
Solve $\left(\frac{1}{9}\right)^{x} = 27$.
- $\frac{1}{9} = 3^{-2}$ and $27 = 3^3$: $3^{-2x} = 3^3$.
- $-2x = 3$.
Respuesta: $x = -\frac{3}{2}$
Not every equation has a common base. $2^x = 10$ can't be written that way, because 10 is not a power of 2. Those equations need logarithms, which come next in What a Logarithm Is and Solving Exponential Equations with Logs.
Error común
Forgetting to multiply exponents when rewriting a base: $8^{x - 1}$ is $(2^3)^{x - 1} = 2^{3x - 3}$, not $2^{3x - 1}$. Put the whole old exponent in parentheses. Check the answer by substituting: $4^3 = 64$ and $8^2 = 64$.