Algebra 2 · F.BF.B.5 · F.LE.A.4

Solving exponential equations by matching bases

The unknown is up in the exponent, where no adding or dividing can reach it. When can you bring it down without a calculator?

You need

Exponential growth and decayGrades 9 to 10 Laws of logarithmsAlgebra 2

Unlocks

Building an exponential model from two pointsPrecalculus

Where you are 11 in Logarithms

Functions13

Logarithms11

▴ 3 above Laws of logarithms Solving by taking a logarithm of both sides

Solving exponential equations by matching bases

Evaluating logarithms, common and natural Logarithmic scales ▾ Show all 11 (3 more)

When you reach for it

An exponential equation has the unknown in an exponent, as in

4^{x + 1} = 8^{x - 2}

Nothing you know from linear equations reaches up there. Subtracting 1 or dividing by 4 does not touch the x, because the x is not a term or a factor; it is a count of how many times 4 is multiplied.

There is one situation where the exponent can be brought down at once, with no calculator: when both sides are powers of the same base. If 2^{\text{something}} = 2^{\text{something else}}, the two somethings must be equal, and that is an ordinary equation. The skill is mostly the rewriting that gets you there: seeing that 4 is 2^2 and 8 is 2^3, that \displaystyle \frac{1}{27} is 3^{-3}, that \displaystyle \sqrt{5} is 5^{1/2}.

Reach for it when the numbers in the equation are all powers of one small number: 2, 4, 8, 16, 32, \displaystyle \frac{1}{2}; or 3, 9, 27, 81, \displaystyle \frac{1}{3}; or 5, 25, 125. When they are not (for example 2^x = 5, since 5 is no whole or fractional power of 2 you can write down), the exponent is a logarithm, and the move is taking a logarithm of both sides.

The procedure

The one fact the method rests on

For a base b with b > 0 and b \neq 1,

b^S = b^T \quad\text{exactly when}\quad S = T

One direction is plain: if S = T then b^S and b^T are the same number. The useful direction is the other one, that two equal powers of the same base must have equal exponents. There are two ways to see it.

From the graph. When b > 1 the function y = b^x rises the whole way, and when 0 < b < 1 it falls the whole way. Either way it never comes back to a height it has already reached, so each output comes from exactly one input: the function is one-to-one. A horizontal line meets the curve at most once.

Figure 1The curve y = 2^x rises the whole way, so it reaches the height 8 at one input only. That is why 2^x = 2^3 forces x = 3.

From the logarithm. Since b^S and b^T are the same positive number, they have the same logarithm base b. By the definition of a logarithm, \log_b\left(b^S\right) = S, because S is exactly the power of b that gives b^S. So

S = \log_b\left(b^S\right) = \log_b\left(b^T\right) = T

This is the same fact read the other way round: \log_b is defined at all only because b^x never takes a value twice, which is why the condition b \neq 1 appears in both places.

The fact also holds when S and T are expressions. If 3^{x - 1} = 3^4, then x - 1 = 4, and since the two equations have exactly the same solutions, solving the second solves the first.

Rewriting both sides as powers of one base

Most equations do not arrive with matching bases. The work is to rewrite each side as a single power of one base, using only the laws of exponents, each of which replaces an expression by an equal one:

Law	Used to rewrite
\left(b^m\right)^n = b^{mn}	8^{x - 2} = \left(2^3\right)^{x - 2} = 2^{3x - 6}
b^m \cdot b^n = b^{m + n}	a product of powers into one power
\dfrac{b^m}{b^n} = b^{m - n}	a quotient of powers into one power
\dfrac{1}{b^n} = b^{-n}	\displaystyle \frac{1}{125} = 5^{-3}
\displaystyle \sqrt[n]{b} = b^{1/n}	\displaystyle \sqrt{5} = 5^{1/2}
b^0 = 1	1 = 7^0

The first law is Powers of powers, products and quotients, the second Multiplying and dividing powers with the same base, the fifth Rational exponents and the last Zero and negative exponents. The first one does most of the work, and it needs brackets: in \left(2^3\right)^{x - 2} the 3 multiplies the whole exponent x - 2, giving 3x - 6, not 3x - 2.

Which base. Choose a number that every base in the equation is a power of. For 4 and 8 that is 2, since 4 = 2^2 and 8 = 2^3; 4 itself will not do, because 8 is not a whole power of 4. For 9 and 27 it is 3. When one choice works and another also does, as 2 or 4 for 16^x = 64, both give the same answer. The smallest whole number that works is the safest choice, because every other base that works is a power of it.

Why every step is safe

Rewriting a side with a law of exponents changes how a number is written, not the number, so the equation keeps exactly its solutions. Matching exponents keeps them too, because the fact above is an “exactly when”. The equation in the exponents is then solved with the moves you already trust.

Because every step can be run backwards, no false answer creeps in. Substituting the answer back is still the best check: work out each side as a power of the same base, and the exponents should agree.

Worked examples

Example 1

One side is already a number

Solve 3^{x - 1} = 81 and 2^{3x} = 64.

Solution

In the first, the left side is a power of 3, so ask what power of 3 gives 81: 3^4 = 81. Then

3^{x - 1} = 3^4

and since the bases match, the exponents are equal: x - 1 = 4, so x = 5. Check: 3^{5 - 1} = 3^4 = 81.

In the second, 64 = 2^6, so 2^{3x} = 2^6 and 3x = 6, giving x = 2. Check: 2^{3 \cdot 2} = 2^6 = 64.

Example 2

Both bases need rewriting

Solve 4^{x + 1} = 8^{x - 2}.

Solution

Neither side is a power of the other’s base, but both 4 and 8 are powers of 2: 4 = 2^2 and 8 = 2^3. Substitute, then multiply exponents by Powers of powers, products and quotients:

\begin{aligned} \left(2^2\right)^{x + 1} &= \left(2^3\right)^{x - 2} \\ 2^{2x + 2} &= 2^{3x - 6} \end{aligned}

The bases match, so the exponents are equal:

2x + 2 = 3x - 6

Subtract 2x from both sides and add 6: x = 8.

Check: the left side is 4^{9} = 2^{18} and the right side is 8^{6} = 2^{18}. Both equal 262{,}144.

Example 3

Fractions and negative exponents

Solve \displaystyle 5^{2x - 1} = \frac{1}{125} and \displaystyle \left(\frac{1}{9}\right)^{x} = 27^{x + 5}.

Solution

For the first, 125 = 5^3, so \displaystyle \frac{1}{125} = \frac{1}{5^3} = 5^{-3}. Then 5^{2x - 1} = 5^{-3}, so 2x - 1 = -3, giving 2x = -2 and x = -1. A negative answer is no problem: \displaystyle 5^{2(-1) - 1} = 5^{-3} = \frac{1}{125}.

For the second, both sides are powers of 3. A fraction with 1 on top is a negative power: \displaystyle \frac{1}{9} = 3^{-2}, and 27 = 3^3. So

\begin{aligned} \left(3^{-2}\right)^{x} &= \left(3^3\right)^{x + 5} \\ 3^{-2x} &= 3^{3x + 15} \end{aligned}

Matching exponents, -2x = 3x + 15, so -5x = 15 and x = -3.

Check: \displaystyle \left(\frac{1}{9}\right)^{-3} = 9^3 = 729, and 27^{-3 + 5} = 27^2 = 729.

Example 4

Roots on one side

Solve \displaystyle 8^{x} = \sqrt{2} and \displaystyle 27^{x - 1} = 9\sqrt{3}.

Solution

A square root is a power with exponent \displaystyle \frac{1}{2}, by Rational exponents: \displaystyle \sqrt{2} = 2^{1/2}. With 8 = 2^3, the first equation becomes

2^{3x} = 2^{1/2}

so \displaystyle 3x = \frac{1}{2} and \displaystyle x = \frac{1}{6}. Check: \displaystyle 8^{1/6} = \left(2^3\right)^{1/6} = 2^{1/2} = \sqrt{2}.

In the second, the right side is a product, so combine it into one power of 3 by Multiplying and dividing powers with the same base: \displaystyle 9\sqrt{3} = 3^2 \cdot 3^{1/2} = 3^{5/2}. The left side is \left(3^3\right)^{x - 1} = 3^{3x - 3}. Matching exponents,

3x - 3 = \frac{5}{2}

so \displaystyle 3x = \frac{11}{2} and \displaystyle x = \frac{11}{6}. Check: \displaystyle 3x - 3 = \frac{11}{2} - 3 = \frac{5}{2}, which is the exponent on the right.

Example 5

A product and a quotient

Solve 5^{x} \cdot 25^{x - 1} = 125 and \dfrac{9^{x + 2}}{3^{x}} = 81.

Solution

The method needs each side to be one power of the base, so a product of powers must be combined first. With 25 = 5^2,

5^{x} \cdot 25^{x - 1} = 5^{x} \cdot 5^{2x - 2} = 5^{3x - 2}

by adding exponents. The right side is 5^3, so 3x - 2 = 3 and \displaystyle x = \frac{5}{3}.

For the quotient, 9^{x + 2} = 3^{2x + 4}, and dividing powers subtracts exponents:

\frac{3^{2x + 4}}{3^{x}} = 3^{(2x + 4) - x} = 3^{x + 4}

The right side is 81 = 3^4, so x + 4 = 4 and x = 0. Check: \dfrac{9^2}{3^0} = \dfrac{81}{1} = 81.

Example 6

An exponent that is quadratic

Solve 2^{x^2} = 4^{x + 4}.

Solution

Rewrite the right side: 4^{x + 4} = 2^{2x + 8}. Matching exponents gives

x^2 = 2x + 8

which is a quadratic equation, not a linear one. Bring everything to one side and factor (see Factoring when the leading coefficient is one):

x^2 - 2x - 8 = 0, \qquad (x - 4)(x + 2) = 0

By the Solving quadratics by factoring, x = 4 or x = -2. Both are solutions, because matching exponents lost nothing and added nothing. Check: for x = 4, 2^{16} against 4^{8} = 2^{16}; for x = -2, 2^{4} = 16 against 4^{2} = 16.

Example 7

How long until a patch of moss covers a wall

A patch of moss on a stone wall covers 500 square centimeters and triples in area every 8 months, so after t months it covers

A = 500 \cdot 3^{t/8}

square centimeters. When does it cover 40{,}500 square centimeters?

Solution

Set A = 40{,}500. The unknown is in the exponent, but there is a factor 500 in the way; the matching-bases fact is about a power standing alone. Divide both sides by 500:

3^{t/8} = \frac{40{,}500}{500} = 81

Now 81 = 3^4, so \displaystyle \frac{t}{8} = 4 and t = 32. The moss covers 40{,}500 square centimeters after 32 months.

That fits the model: 32 months is 4 tripling periods, and 500 \cdot 3^4 = 500 \cdot 81 = 40{,}500. This is the same reasoning as counting growth periods, written as an equation.

Example 8

A quadratic hiding in the powers

Solve 9^{x} - 8 \cdot 3^{x} - 9 = 0.

Solution

There are three terms, so the two sides cannot both be single powers as they stand. But 9^{x} = \left(3^2\right)^{x} = \left(3^{x}\right)^2, so the equation is a quadratic in the quantity 3^{x}. Write u = 3^{x}:

u^2 - 8u - 9 = 0, \qquad (u - 9)(u + 1) = 0

so u = 9 or u = -1. Now go back to x. The first gives 3^{x} = 9 = 3^2, so x = 2. The second asks for 3^{x} = -1, which has no solution: a positive base raised to any real power is positive. So the only solution is x = 2.

Check: 9^2 - 8 \cdot 3^2 - 9 = 81 - 72 - 9 = 0.

Boundaries

The base must be positive and not 1. With base 1, every power is 1: 1^{2} = 1^{7} although 2 \neq 7, and 1^{x} = 1 holds for every x. With base 0, 0^{2} = 0^{5}. With a negative base, (-2)^2 = (-2)^4 is false but (-1)^2 = (-1)^4 is true, and (-2)^{x} is not even a real number for \displaystyle x = \frac{1}{2}. The fact b^S = b^T \Rightarrow S = T is stated only for b > 0, b \neq 1, and only there is it true.

Each side must be a single power of the base. 3 \cdot 2^{x} = 24 must first be divided by 3, giving 2^{x} = 8. A sum such as 2^{x} + 2^{x + 1} is not a power until it has been factored, 2^{x}(1 + 2) = 3 \cdot 2^{x}. Matching the exponents of the terms of a sum one by one is not an allowed step.

The numbers must share a base you can write down. In 2^{x} = 5 there is a solution, a little over 2, but 5 is not a power of 2 with an exponent you can name in advance. Its exponent is \log_2 5, and finding it is the job of Solving by taking a logarithm of both sides and Change of base.

Some equations have no solution. For b > 0, b^x is positive for every real x, so 3^{x} = -9 and 2^{x} = 0 have no real solution. Writing -9 as a power of 3 is impossible, not just awkward.

Real exponents only. Everything here treats x as a real number, which is where the fact is proved. Over the complex numbers b^x is no longer one-to-one, and an equation such as 2^x = 8 has other, non-real solutions.

Where this goes next

The same fact solves for the factor in a model. Knowing two values of y = a \cdot b^{x} and dividing one equation by the other leaves a single power of b, which is how an exponential model is built from two points.

When the bases do not match, logarithms take over: Solving by taking a logarithm of both sides brings any exponent down, and Change of base lets a calculator finish the job. The matching-bases move is the special case where that logarithm happens to come out as a number you can write by hand.

Logarithms have the mirror fact. Because \log_b is one-to-one on the positive numbers, \log_b M = \log_b N forces M = N, for M > 0 and N > 0. That is how logarithmic equations are solved, with the extra care that an answer must keep every logarithm’s input positive.

Exercises

One side is a number

Solve 5^{x - 2} = 125.
Solve 3^{2x + 1} = 243.
Solve 7^{4x} = 7^{x + 9}.
Solve 10^{x} = 0.001.

Rewrite both sides

Solve 9^{x} = 27.
Solve 16^{x - 1} = 8^{x}.
Solve 25^{x + 3} = 125^{x}.
Solve 27^{2x} = 9^{x + 4}.
Solve 32^{x} = 4^{x - 3}.

Fractions, negatives and roots

Solve \displaystyle \left(\frac{1}{2}\right)^{x} = 16.
Solve \displaystyle 4^{x} = \frac{1}{32}.
Solve \displaystyle \left(\frac{1}{25}\right)^{x - 1} = 125.
Solve \displaystyle 9^{x} = \sqrt{3}.
Solve \displaystyle 8^{x + 1} = 4\sqrt{2}.
Solve \displaystyle \left(\frac{2}{3}\right)^{x} = \frac{9}{4}.

Products, quotients and quadratics

Solve 3^{x} \cdot 9^{x + 1} = 27^{3}.
Solve \dfrac{2^{5x}}{2^{x + 2}} = 64.
Solve 5^{x^2 - x} = 15{,}625.
Solve 6 \cdot 3^{x} = 162.
Solve 2^{x} + 2^{x + 3} = 72.
Solve 4^{x} - 10 \cdot 2^{x} + 16 = 0.

Problems

A breeding tank holds 200 crickets, and the number triples every 5 days, so after t days there are C = 200 \cdot 3^{t/5} crickets. After how many days are there 48{,}600?
Each layer of a water filter lets through \displaystyle \frac{1}{3} of the dye that reaches it. With n layers, the fraction of dye that gets through is \displaystyle \left(\frac{1}{3}\right)^{n}. How many layers let through \displaystyle \frac{1}{243} of the dye?

Reasoning

Someone writes: “1^{x} = 1^{3}, the bases match, so x = 3.” Explain what is wrong, and give the true solution set.
The bases in 5^{x} = 7^{x} do not match. Show that the equation still has exactly one solution, and find it, by dividing both sides by 7^{x}.
(-2)^3 = -8, yet 2^{x} = -8 has no real solution. Explain why, and why the method cannot use -2 as a base instead.
Using the logarithm \log_b, prove that b^S = b^T implies S = T when b > 0 and b \neq 1. Say where the condition b \neq 1 is needed.

Answers

x = 5. 125 = 5^3, so x - 2 = 3.
x = 2. 243 = 3^5, so 2x + 1 = 5.
x = 3. The bases already match: 4x = x + 9, so 3x = 9.
x = -3. \displaystyle 0.001 = \frac{1}{1000} = 10^{-3}.
\displaystyle x = \frac{3}{2}. 9^{x} = 3^{2x} and 27 = 3^3, so 2x = 3.
x = 4. In base 2: 2^{4x - 4} = 2^{3x}, so 4x - 4 = 3x.
x = 6. In base 5: 5^{2x + 6} = 5^{3x}, so 2x + 6 = 3x.
x = 2. In base 3: 3^{6x} = 3^{2x + 8}, so 4x = 8.
x = -2. In base 2: 2^{5x} = 2^{2x - 6}, so 3x = -6.
x = -4. \displaystyle \frac{1}{2} = 2^{-1}, so 2^{-x} = 2^4 and -x = 4.
\displaystyle x = -\frac{5}{2}. 2^{2x} = 2^{-5}, so 2x = -5.
\displaystyle x = -\frac{1}{2}. \displaystyle \frac{1}{25} = 5^{-2}, so 5^{-2x + 2} = 5^3, -2x + 2 = 3 and -2x = 1.
\displaystyle x = \frac{1}{4}. 3^{2x} = 3^{1/2}, so \displaystyle 2x = \frac{1}{2}.
\displaystyle x = -\frac{1}{6}. \displaystyle 4\sqrt{2} = 2^2 \cdot 2^{1/2} = 2^{5/2} and 8^{x + 1} = 2^{3x + 3}, so \displaystyle 3x + 3 = \frac{5}{2} and \displaystyle 3x = -\frac{1}{2}.
x = -2. \displaystyle \frac{9}{4} = \left(\frac{3}{2}\right)^2 = \left(\frac{2}{3}\right)^{-2}, so the bases match already.
\displaystyle x = \frac{7}{3}. The left side is 3^{x} \cdot 3^{2x + 2} = 3^{3x + 2} and the right is 3^{9}, so 3x + 2 = 9.
x = 2. The quotient is 2^{5x - (x + 2)} = 2^{4x - 2} and 64 = 2^6, so 4x - 2 = 6.
x = 3 or x = -2. 15{,}625 = 5^6, so x^2 - x = 6, (x - 3)(x + 2) = 0.
x = 3. Divide by 6 first: 3^{x} = 27 = 3^3.
x = 3. Factor: 2^{x} + 2^{x} \cdot 2^3 = 2^{x}(1 + 8) = 9 \cdot 2^{x}, so 2^{x} = 8 = 2^3.
x = 1 or x = 3. With u = 2^{x}, u^2 - 10u + 16 = 0, (u - 2)(u - 8) = 0, so 2^{x} = 2 or 2^{x} = 8.
After 25 days. \displaystyle 3^{t/5} = \frac{48{,}600}{200} = 243 = 3^5, so \displaystyle \frac{t}{5} = 5.
5 layers. \displaystyle \frac{1}{243} = \left(\frac{1}{3}\right)^5, so n = 5.
Base 1 is excluded: 1^{x} = 1 for every x, so 1^x is not one-to-one and equal powers say nothing about the exponents. The equation reads 1 = 1, and every real number is a solution.
x = 0, the only solution. 7^{x} > 0, so dividing is allowed and gives \displaystyle \left(\frac{5}{7}\right)^{x} = 1 = \left(\frac{5}{7}\right)^{0}. The base \displaystyle \frac{5}{7} is positive and not 1, so the exponents match: x = 0.
2^{x} is positive for every real x, so it never equals -8. Writing -8 = (-2)^3 does not help, because the left side is 2^{x}, not (-2)^{x}, and the two differ (at x = 1 they are 2 and -2). A negative base is excluded from the method in any case: (-2)^{x} is not a real number at \displaystyle x = \frac{1}{2}, and (-1)^2 = (-1)^4 shows equal powers of a negative base need not have equal exponents.
b^S and b^T are the same positive number, so they have the same logarithm base b. By the definition of the logarithm, \log_b\left(b^S\right) = S and \log_b\left(b^T\right) = T, so S = T. The condition b \neq 1 is what makes \log_b exist: the definition needs each positive number to be exactly one power of b, and for b = 1 the only power is 1, reached by every exponent.