Exponents and Logarithms

How to solve exponential equations with the same base?

Quick answer

When an exponential equation shares the same base on both sides, set the exponents equal to each other using the one-to-one property and solve for the variable. If terms share a base across sums, factor out the lowest power. If expressions feature doubled exponents, apply substitution to convert the equation into a solvable quadratic.

The topic explained

The basic rule: matching bases

Solving exponential equations that share a common base relies on the one-to-one property. If b to the power of M equals b to the power of N (where the base b is positive and not equal to 1), then M must equal N. Once both sides of an equation are expressed with identical bases, you can drop the bases and equate the exponents directly.

For example, consider 2^(x + 3) = 16. Recognize 16 as 2^4, so the equation becomes 2^(x + 3) = 2^4. Setting the exponents equal gives x + 3 = 4, which yields x = 1. Always look for powers of familiar integers such as 2, 3, 5, or 10.

Factoring out a common power

Sometimes an equation contains multiple terms with the same base added or subtracted, such as 3^(x + 2) - 3^x = 72. You cannot simply equate exponents across addition. Instead, use exponent laws in reverse to separate the powers: write 3^(x + 2) as 3^x * 3^2, which is 9 * 3^x.

Now factor out the shared term 3^x: 3^x * (9 - 1) = 72, which simplifies to 3^x * 8 = 72. Dividing both sides by 8 gives 3^x = 9. Since 9 is 3^2, equate the exponents to find x = 2.

Quadratic form and extraneous roots

More advanced problems resemble quadratics. An equation like 4^x - 3 * 2^x - 4 = 0 can be rewritten using base 2 as (2^x)^2 - 3 * (2^x) - 4 = 0. Let u = 2^x to produce the quadratic u^2 - 3u - 4 = 0, which factors into (u - 4)(u + 1) = 0.

This gives candidate values u = 4 and u = -1. Substitute back: 2^x = 4 gives x = 2. However, 2^x = -1 has no real solution because a positive base raised to any real power is strictly positive. Always reject negative or zero values when solving for u.

Free printable worksheet

Solving Exponential Equations with a Common Base

Solve each exponential equation for x by expressing both sides in a common base, factoring, or using quadratic substitution. Check for extraneous roots.

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Solving Exponential Equations with a Common BaseName:

Solve: 5^(2x - 1) = 125
Solve: 4^(x + 1) = 8^(x - 1)
Solve: (1/3)^x = 27^(x + 2)
Solve: 2^(x + 3) + 2^x = 36
Solve: 9^x - 4 * 3^x + 3 = 0
Solve: 4^x - 5 * 2^x - 24 = 0

Common questions

What grade level covers this method?

This topic is taught in Grade 11 (ages 16 to 17) as part of Algebra 2 or Precalculus, establishing essential algebra skills before students introduce logarithms.

What if the two sides cannot be written with the same base?

When terms cannot be converted into a common integer or rational base, such as 2^x = 5, you must use logarithms rather than equating exponents.

Why must negative values of u be rejected during substitution?

If you substitute u = b^x where b is a positive base, b^x is strictly positive for every real number x. Therefore, u can never equal zero or a negative number.

What is the most common mistake students make?

Students often multiply bases and exponents together or distribute powers incorrectly across addition, such as writing 2^x + 2^x as 4^x instead of 2 * 2^x = 2^(x + 1).

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