Solving Exponential Equations with Logarithms

When we face an exponential equation of the form $$ b^x = c $$ a very effective strategy is to apply a logarithm to both sides. Using any base works, and the key idea is simple: $$ \log_b(b^x) = \log_b(c) $$ This immediately brings the exponent down and gives $$ x = \log_b(c) $$

This approach is widely used in algebra and calculus. It works reliably as long as two conditions are met:

• both sides of the equation must be positive, since logarithms are defined only for positive numbers,
• the argument of the logarithm should not involve sums or differences, because logarithmic identities cannot simplify addition or subtraction.

Example. A term like $$ \log(1 + 4x) $$ cannot be simplified using logarithmic properties. A product inside the logarithm, however, can be handled using the product rule.

A worked example

Let's look at a concrete case:

$$ 5^x = 9 $$

This equation is ideal for logarithmic methods because both sides are positive. We can take the logarithm of each side using any base. Choosing base 5 keeps the expressions clean:

$$ \log_5(5^x) = \log_5(9) $$

Using the power rule we bring the exponent down:

$$ x \cdot \log_5(5) = \log_5(9) $$

Now we isolate the variable by dividing both sides by \( \log_5(5) \):

$$ \frac{x \cdot \log_5(5)}{\log_5(5)} = \frac{\log_5(9)}{\log_5(5)} $$

$$ x = \frac{\log_5(9)}{\log_5(5)} $$

Since \( \log_5(5) = 1 \), the expression becomes simply:

$$ x = \log_5(9) $$

This is the exact solution of the exponential equation. It shows how logarithms turn what looks like a difficult exponent problem into a straightforward algebraic step.

And the same method applies to many other exponential equations.