How To Take Log Of Both Sides

Ever stared at a math problem involving an exponent and felt that immediate sense of "I don't know how to get that variable down from there"? It’s a common wall to hit. You have an equation like $5^x = 50$, and suddenly, the $x$ is floating somewhere in the stratosphere, out of reach and impossible to isolate using basic addition or subtraction.

This is where logarithms come in. So if you want to solve for an unknown power, you need to learn how to take the log of both sides. But they are essentially the "undo" button for exponents. It sounds like a technical chore, but once you grasp the logic, it becomes one of the most powerful tools in your algebraic toolkit.

What Is Taking the Log of Both Sides

At its core, taking the log of both sides is an application of the golden rule of algebra: whatever you do to one side of an equation, you must do to the other. If $A = B$, then $\log(A) = \log(B)$.

But why do we do this specifically? Because logarithms have a unique property that turns multiplication into addition and, more importantly, turns exponentiation into multiplication.

The Power Rule

The real magic happens because of the power rule* of logarithms. This rule states that $\log(x^n) = n \cdot \log(x)$. This is the "secret sauce." It allows you to take that pesky exponent—the thing that is making the equation difficult—and pull it down to the ground level where it becomes a simple multiplier.

Which Log Should You Use?

You’ll usually see two types of logs in your textbooks. The first is the common logarithm (log base 10), which is often just written as $\log$. The second is the natural logarithm (ln), which uses the mathematical constant $e$ as its base.

In most algebra problems, it doesn't matter which one you pick, as long as you are consistent. If you take the common log of the left side, you must take the common log of the right side. If you use natural logs, use them on both sides. The math will work out the same way in the end.

Why It Matters

Why should you care about this beyond passing a calculus exam? Because growth is rarely linear.

In the real world, things like population growth, compound interest in a bank account, or the spread of a virus don't move in straight lines. Consider this: they move exponentially. If you are trying to figure out how many years it will take for your savings to triple, or how long it will take for a bacterial culture to reach a certain density, you are dealing with exponents.

Without the ability to take the log of both sides, you're stuck guessing and checking numbers. Taking the log of both sides gives you a direct, mathematical path to the exact answer. Because of that, you'd be sitting there typing "2. No. Also, 5 years? Because of that, " into a calculator, hoping to get close. Because of that, no. But 3 years? It turns a "guessing game" into a "solving game.

How to Take the Log of Both Sides

Let’s break this down into a repeatable process. I find that most people struggle not because they don't understand the concept, but because they get lost in the algebraic "mess" that happens halfway through.

Step 1: Isolate the Exponential Term

Before you even touch a logarithm, look at your equation. Is there a coefficient hanging out in front of your base? As an example, if you have $3 \cdot 2^x = 50$, you shouldn't take the log yet.

If you take the log of $3 \cdot 2^x$ right now, you'll end up with $\log(3 \cdot 2^x)$, which requires the product rule* to expand. Consider this: it's much cleaner to divide both sides by 3 first. Worth adding: get your equation into the form $Base^{exponent} = Number$. Once it looks like that, you're ready.

Step 2: Apply the Logarithm to Both Sides

Once you have $2^x = 16.66$ (which is $50/3$), you apply the log. You can write it as: $\log(2^x) = \log(16.66)$

At this stage, it looks like you've just made the equation look more complicated. Here's the thing — you haven't. You've actually set the stage for the most important step.

Step 3: Use the Power Rule to "Drop" the Exponent

This is the moment of truth. Use that power rule we talked about earlier. The $x$ that was stuck up in the exponent can now be moved to the front as a multiplier. Not complicated — just consistent.

$x \cdot \log(2) = \log(16.66)$

Now, look at that. Think about it: the $x$ is no longer an exponent. It's just a variable being multiplied by a constant (since $\log(2)$ is just a number you can find on a calculator).

Step 4: Solve for the Variable

This is the easiest part. Since $x$ is being multiplied by $\log(2)$, you simply divide both sides by $\log(2)$ to isolate $x$.

$x = \frac{\log(16.66)}{\log(2)}$

Plug that into your calculator, and you have your answer. No guessing required.

Common Mistakes / What Most People Get Wrong

I've seen students (and even some professionals) trip over the same hurdles repeatedly. If you want to avoid frustration, watch out for these.

For more on this topic, read our article on the tortoise and the hare story or check out what is the area of the triangle shown below.

The "Log of the Sum" Trap This is the biggest mistake. You cannot take the log of a sum and have it behave nicely. If you have $2^x + 5 = 20$, you cannot say $\log(2^x + 5) = \log(20)$. And you definitely cannot say $\log(2^x) + \log(5) = \log(20)$.

Logarithms distribute over multiplication and division, but they do not distribute over addition or subtraction. If you see a plus or minus sign, you must use standard algebra to isolate the exponential term before* you apply the log.

Forgetting the Base While you can use any base, if your problem specifically uses $e$ (the natural base), it is much easier to use the natural log ($\ln$). If you mix them up—using $\log$ on one side and $\ln$ on the other—you'll get a completely wrong answer.

Misplacing the Coefficient Going back to my earlier example: $3 \cdot 2^x = 50$. If you take the log of both sides immediately, you get $\log(3 \cdot 2^x) = \log(50)$. To solve this, you'd have to use the product rule: $\log(3) + \log(2^x) = \log(50)$. Then you'd have to move $\log(3)$ to the other side. It works, but it's a lot of extra room for error. Always simplify the "outside" of the exponent first.

Practical Tips / What Actually Works

If you want to master this, stop looking at it as a series of abstract rules and start looking at it as a workflow.

Always check your work with a calculator. Since these problems often involve decimals, it's easy to make a rounding error. Once you get your value for $x$, plug it back into the original equation ($2^x$). If it doesn't equal your target number, you missed a step.
Use the Natural Log ($\ln$) for Calculus. If you are moving into higher-level math or physics, you will spend almost all your time using $\ln$. It's a good habit to get used to it now.
Keep your decimals long during the process. If you round $\log(2)$ to just "0.3" early in the problem, your final answer will be significantly off. Keep as many digits as possible until the very last step.
Recognize the pattern. Whenever you see a variable in a power, your brain should immediately think: "Logarithms." It'

the go-to tool for pulling that variable down into the realm of basic algebra.

Putting It All Together

Let’s walk through a complete example using everything we’ve covered.

Example Problem:
Solve for $x$ in the equation:
$4 \cdot 3^{2x + 1} = 100$

Step-by-step solution:

Isolate the exponential term.
Divide both sides by 4 to get the term with $x$ alone on one side:
$3^{2x + 1} = \frac{100}{4} = 25$
Apply the natural logarithm to both sides.
Take $\ln$ of both sides (we’ll use $\ln$ here since it’s standard for advanced work):
$\ln(3^{2x + 1}) = \ln(25)$
Use the power rule to bring the exponent down.
The exponent $(2x + 1)$ can be moved in front of the log:
$(2x + 1)\ln(3) = \ln(25)$
Solve for $x$ using algebra.
Divide both sides by $\ln(3)$:
$2x + 1 = \frac{\ln(25)}{\ln(3)}$

Subtract 1 from both sides:
$2x = \frac{\ln(25)}{\ln(3)} - 1$

Finally, divide by 2:
$x = \frac{1}{2}\left(\frac{\ln(25)}{\ln(3)} - 1\right)$
Calculate the numerical value.
Using a calculator:
- $\ln(25) \approx 3.2189$
- $\ln(3) \approx 1.0986$
  So,
  $x = \frac{1}{2}\left(\frac{3.2189}{1.0986} - 1\right) = \frac{1}{2}(2.9300 - 1) = \frac{1}{2}(1.9300) = 0.965$
Check your answer.
Plug $x = 0.965$ back into the original equation:
$4 \cdot 3^{2(0.965) + 1} = 4 \cdot 3^{2.93} \approx 4 \cdot 24.99 = 99.96$
This is very close to 100, confirming our solution is correct.

Conclusion

Solving exponential equations doesn’t have to be mysterious or reliant on guesswork. Remember to avoid common pitfalls like misapplying logarithm rules to sums or mixing up log bases, and always verify your answer by plugging it back into the original equation. By following a clear, systematic approach—isolating the exponential term, applying logarithms, using the power rule, and solving algebraically—you can tackle any problem of this type with confidence. With practice, recognizing when to deploy logarithms becomes second nature, turning what once seemed like an impossible puzzle into a straightforward calculation.