To solve for a variable trapped in an exponent, you use logarithms to bring it down to a normal position where standard algebra works. The core idea: taking the logarithm of both sides lets you convert an exponent into a multiplier, turning a problem like 2^x = 48 into x = log₂(48). There are two main approaches depending on the equation, and choosing the right one saves significant time.
Method 1: Rewrite Both Sides With the Same Base
When both sides of an equation can be expressed as powers of the same number, you can skip logarithms entirely. The rule is simple: if b^x = b^y, then x = y. Once the bases match, the exponents must be equal, and you solve what’s left as a basic algebra problem.
Take 16^x = 64. Both 16 and 64 are powers of 2: 16 is 2⁴ and 64 is 2⁶. Rewriting gives you (2⁴)^x = 2⁶. When you raise a power to a power, you multiply the exponents, so this becomes 2^(4x) = 2⁶. Now just set the exponents equal: 4x = 6, and x = 3/2.
This method works cleanly when you’re dealing with numbers like 2, 4, 8, 16, 32, 64 (all powers of 2), or 3, 9, 27, 81 (powers of 3), or 5, 25, 125, and so on. It’s also the primary technique tested on the SAT, where the expectation is that you recognize shared bases. The exponent rules you need to remember: when multiplying same bases, add exponents; when dividing same bases, subtract exponents; when raising a power to a power, multiply exponents.
Method 2: Use Logarithms
Most exponential equations don’t have bases you can match up neatly. When you’re facing something like 5 · 2^x = 240, logarithms are the tool that pulls the variable out of the exponent. Here’s the process, broken into clear steps.
Step 1: Isolate the Exponential Term
Before touching logarithms, get the exponential expression alone on one side of the equation with a coefficient of 1. This means moving everything else away using basic operations. For 5 · 2^x = 240, divide both sides by 5 first to get 2^x = 48. For an equation like 7 + 15e^(1−3z) = 10, subtract 7, then divide by 15, arriving at e^(1−3z) = 1/5. Skipping this step is one of the most common mistakes, and it makes the logarithm work far messier than it needs to be.
Step 2: Take the Logarithm of Both Sides
Once the exponential is isolated, apply a logarithm to both sides. You can use any logarithm base and get the right answer, but two choices make life easier. Use the natural log (ln) when the base of your exponential is e, since ln(e^something) simplifies to just the “something.” Use the common log (log base 10) when your base is 10, for the same reason. For any other base, either log or ln works fine on a calculator.
Taking ln of both sides of e^(1−3z) = 1/5 gives you ln(e^(1−3z)) = ln(1/5), which simplifies immediately to 1 − 3z = ln(1/5).
Step 3: Apply the Power Rule to Bring Down the Exponent
This is the key property that makes the whole method work. The power rule of logarithms states that log(M^p) = p · log(M). In plain terms, an exponent inside a logarithm can move out front as a multiplier. That’s how the variable escapes from the exponent position and lands somewhere you can solve for it with regular algebra.
For 2^x = 48, taking log of both sides gives log(2^x) = log(48). The power rule converts the left side: x · log(2) = log(48). Now x is just being multiplied by a number, so divide both sides by log(2) to get x = log(48)/log(2). Plug that into a calculator and you get approximately 5.585.
Step 4: Solve the Remaining Algebra
After the exponent comes down, you’re left with a standard linear equation. For the equation 6 · 10^(2x) = 48, isolating gives 10^(2x) = 8. Converting to logarithmic form: log₁₀(8) = 2x. Then x = log(8)/2, which is approximately 0.4515. If the exponent contained an expression like (1 − 3z), you’d solve 1 − 3z = ln(1/5) by subtracting 1 and dividing by −3.
When Variables Appear on Both Sides
Equations like 3^x = 7^(x−2) have the variable in exponents on both sides. Take the log of both sides, then use the power rule on each: x · log(3) = (x − 2) · log(7). Now distribute on the right: x · log(3) = x · log(7) − 2 · log(7). Get all terms with x on one side: x · log(3) − x · log(7) = −2 · log(7). Factor out x: x(log 3 − log 7) = −2 · log(7). Divide to isolate x: x = −2 · log(7) / (log 3 − log 7).
The key here is treating log(3) and log(7) as ordinary numbers (which they are) and using the distributive property and factoring just as you would with any linear equation.
Exponential Equations in Quadratic Form
Some equations, like e^(2x) − 5e^x + 6 = 0, look like quadratic equations in disguise. The trick is substitution: let u = e^x, which makes e^(2x) = u². The equation becomes u² − 5u + 6 = 0, which factors into (u − 2)(u − 3) = 0. So u = 2 or u = 3, meaning e^x = 2 or e^x = 3. Taking ln of each gives x = ln(2) or x = ln(3).
Always check your substitution results before solving. If the factoring had produced something like e^x = −7, you’d throw that solution out. A positive number raised to any real power never equals a negative number, so there’s no value of x that works. These are called extraneous solutions, and they appear specifically in quadratic-form exponential equations.
Common Mistakes to Avoid
- Trying to “distribute” a logarithm across addition. log(a + b) does NOT equal log(a) + log(b). Logarithms only split across multiplication (log(ab) = log a + log b) and division. This is why isolating the exponential term first is so important.
- Forgetting that log arguments must be positive. You can only take the logarithm of a positive number. If your algebra leads to logging a negative number or zero, the equation has no solution at that step.
- Not isolating before taking logs. Applying ln to both sides of 7 + 15e^(1−3z) = 10 without first simplifying creates a mess that’s nearly impossible to untangle. Always move constants and coefficients away from the exponential term first.
- Confusing when exponent rules apply. The rules for combining exponents only work when bases are the same and terms are being multiplied or divided. They do not apply to 2³ + 2⁴ or 2³ · 5⁴.
Choosing the Right Log on a Calculator
Your calculator has two log buttons: “log” (base 10) and “ln” (base e, approximately 2.718). For solving exponential equations, it genuinely does not matter which one you pick. Using log base 10 on both sides or natural log on both sides will produce the same final answer, because the choice of base cancels out during division. The expression log(48)/log(2) equals ln(48)/ln(2) equals approximately 5.585 either way.
The one efficiency shortcut: if the equation involves e as the base, using ln eliminates a step because ln(e^something) collapses directly to the exponent. Similarly, if the base is 10, the common log button does the same thing. For every other base, you’ll end up dividing one log by another regardless of which button you use. This division is essentially the change of base formula: log_b(x) = log(x) / log(b), which converts any logarithm base into one your calculator can handle.