What if there are exponential terms on both sides with different bases?

Take the logarithm of both sides immediately, apply the power rule to bring down both exponents, expand the expressions, and group all terms containing the variable on one side to factor it out.

Which logarithm base should I use?

You can use any valid base. The natural logarithm ($$\ln$$) is standard and simplifies equations containing base e. Base 10 ($$\log$$) is also common. The final numerical answer will be identical regardless of the base chosen.

What if the isolated exponential equals a negative number?

The equation has no real solution. A positive base raised to a real power is always positive, so an equation like $$2^x = -4$$ cannot be solved in the real number system.

How to solve an exponential equation using logarithms

Solving an exponential equation requires isolating the exponential term and taking the logarithm of both sides to bring the variable out of the exponent. This method applies when the variable is in the exponent and the bases cannot be easily matched to use the one-to-one property.

The setup

Isolate the term with the variable in the exponent on one side of the equation. Ensure the other side of the equation is strictly positive, as the logarithm of a negative number or zero is undefined in the real number system.

The steps

Isolate the exponential expression. 2. Take the natural logarithm ( $\ln$ ) or common logarithm ( $\log$ ) of both sides. 3. Use the power rule of logarithms, $\ln(a^x) = x \ln(a)$ , to bring the exponent down as a multiplier. 4. Solve the resulting linear or algebraic equation for the variable.

Checking the result

Substitute the numerical answer back into the original equation. Evaluate both sides to verify they result in the same value, accounting for minor discrepancies due to rounding errors if using a decimal approximation.

Common errors

A frequent mistake is failing to isolate the exponential term before taking the logarithm. Another is attempting to distribute a logarithm across addition, incorrectly assuming $\ln(x+y) = \ln(x) + \ln(y)$ . Finally, taking the logarithm of a negative constant indicates no real solution exists, which is often mistakenly ignored.

Worked example

Solve $3e^{2x - 1} + 4 = 19$ for x.

Subtract 4 from both sides: $3e^{2x - 1} = 15$ . Divide both sides by 3: $e^{2x - 1} = 5$ . Take the natural logarithm of both sides: $\ln(e^{2x - 1}) = \ln(5)$ . Apply the power rule and the identity $\ln(e) = 1$ : $2x - 1 = \ln(5)$ . Add 1 to both sides: $2x = \ln(5) + 1$ . Divide by 2: $x = \frac{\ln(5) + 1}{2}$ . The exact solution is $x = \frac{\ln(5) + 1}{2}$ , which evaluates to approximately 1.3047.

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References: OpenStax College Algebra, Chapter 6: Exponential and Logarithmic Functions · Khan Academy: Exponential equations with logarithms