How to Solve an Exponential Equation with Different Bases

Solving an exponential equation with different bases requires either rewriting the expressions to share a common base or using logarithms to isolate the variable. When the bases cannot be made the same, logarithms provide a reliable method. This skill is used in algebra and models real-world processes like population growth and decay.

Direct Method: When Bases Can Be Rewritten

First, check if the bases are powers of the same number. For example, 4 and 8 are both powers of 2: $4 = 2^2$, $8 = 2^3$. So $4^x = 8^{x-1}$ becomes $(2^2)^x = (2^3)^{x-1}$, or $2^{2x} = 2^{3x-3}$. Since the bases are now equal, set the exponents equal: $2x = 3x - 3$, so $x = 3$.

Other common conversions:

• $9 = 3^2$
• $27 = 3^3$
• $16 = 2^4 = 4^2$
• $32 = 2^5$

If both sides can be expressed with the same base, equate the exponents and solve. This avoids logarithms entirely. Always try this first Khan Academy.

Using Logarithms When Bases Differ

When no common base exists, use logarithms. The power rule $\log(a^x) = x \log(a)$ allows you to bring the exponent down, turning the equation into a linear one.

Solve $3^x = 7$.
Take the logarithm of both sides: $\log(3^x) = \log(7)$.
Apply the power rule: $x \log(3) = \log(7)$.
Then $x = \frac{\log(7)}{\log(3)}$.

Using approximate values, $\log(7) \approx 0.845$ and $\log(3) \approx 0.477$, so $x \approx \frac{0.845}{0.477} \approx 1.77$.
Using natural logs: $x = \frac{\ln(7)}{\ln(3)} \approx \frac{1.946}{1.099} \approx 1.77$, the same result.

The choice of log—common or natural—does not change the answer. Just be consistent Paul’s Online Math Notes.

One Worked Example

Solve $4^{2x} = 9$.

Note that $4 = 2^2$ and $9 = 3^2$, but 2 and 3 share no common base. So logarithms are needed.

Take the log of both sides: $\log(4^{2x}) = \log(9)$.
Apply the power rule: $2x \log(4) = \log(9)$.
Then $x = \frac{\log(9)}{2 \log(4)}$.

Now simplify. Since $9 = 3^2$, $\log(9) = 2 \log(3)$.
And $\log(4) = \log(2^2) = 2 \log(2)$.
So $x = \frac{2 \log(3)}{2 \cdot 2 \log(2)} = \frac{\log(3)}{2 \log(2)}$.

Using approximations: $\log(3) \approx 0.477$, $\log(2) \approx 0.301$, so
$x \approx \frac{0.477}{2 \cdot 0.301} = \frac{0.477}{0.602} \approx 0.792$.

Check: $4^{2(0.792)} = 4^{1.584}$. Since $4^{1.5} = (2^2)^{1.5} = 2^3 = 8$, and $4^{1.584}$ is slightly higher, close to 9. The solution is reasonable.

Substitution for Quadratic-Like Forms

Some equations resemble quadratics. For example: $3^{2x} - 4 \cdot 3^x + 3 = 0$.

Let $u = 3^x$, so $3^{2x} = u^2$. The equation becomes $u^2 - 4u + 3 = 0$.
Factoring: $(u - 3)(u - 1) = 0$, so $u = 3$ or $u = 1$.
Then $3^x = 3$ gives $x = 1$, and $3^x = 1$ gives $x = 0$.

Both satisfy the original equation. Substitution simplifies complex forms but always verify solutions.

Checking Solutions

Exponential expressions are always positive. If a step leads to $a^x = -2$, discard it—no real solution exists. After substitution or algebraic manipulation, plug solutions back in.

For example, solving $4^x + 2^x = 6$:
Let $u = 2^x$, so $4^x = (2^x)^2 = u^2$.
Then $u^2 + u - 6 = 0$, which factors as $(u + 3)(u - 2) = 0$.
So $u = -3$ or $u = 2$. But $2^x = -3$ has no real solution. Only $2^x = 2$ gives $x = 1$.
Check: $4^1 + 2^1 = 4 + 2 = 6$. Valid.

Change of Base Formula

The formula $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$ lets you compute logarithms in any base using common or natural logs. For example, solving $3^x = 7$ gives $x = \log_3(7) = \frac{\log(7)}{\log(3)}$, which matches earlier steps.

This ensures consistency across methods and tools, especially when calculators use only base 10 or base $e$ Khan Academy.

Key Takeaways

• First, check if different bases can be rewritten as powers of the same number. If so, equate the exponents.
• When no common base exists, take the logarithm of both sides and apply the power rule.
• For equations with terms like $a^{2x}$ and $a^x$, use substitution to form a quadratic.
• Always verify solutions by substituting back, as some steps may introduce invalid results.
• The change of base formula allows flexibility in computation and confirms consistency across log types.