Solving Exponential and Logarithmic Equations

Learning goals

Use one-to-oneness to equate exponents on a common base
Take a logarithm to bring an unknown exponent down
Turn $\log_b u = k$ into $u = b^k$
Check every candidate when combining logs enlarges the domain
Reject a substitution value $u \le 0$ , which yields no $x$
State a rounded answer as rounded, keeping the exact form

One-to-one is the license

Return to the equation from the box above, $2^x = 32$ . Since $32 = 2^5$ , the equation is really $2^x = 2^5$ . The exponents must match because the bases already do, so $x = 5$ . That move, matching exponents because the bases match, is worth naming precisely, because every technique in this lesson is some version of it.

A function $f$ is one-to-one when different inputs always give different outputs, or equivalently, when $f(u) = f(v)$ forces $u = v$ . On a graph this says that no horizontal line meets the curve twice. That is precisely the condition you met in Inverse Functions: a function has an inverse exactly when it is one-to-one. The reason is that the inverse has to send each output back to a single input.

Fix a base $b$ with $b > 0$ and $b \neq 1$ . The logarithm $\log_b$ was defined as the inverse of $f(x) = b^x$ , so the two of them cancel:

\log_b\!\left(b^t\right) = t \quad \text{for every real } t, \qquad\qquad b^{\log_b M} = M \quad \text{for every } M > 0.

Read them in words. The first says: the exponent you must put on $b$ to get $b^t$ is $t$ . The second says: if you put the exponent $\log_b M$ on $b$ , you get $M$ . Everything below is squeezed out of these two identities.

Equating exponents and equating arguments#

Take a base $b$ with $b > 0$ and $b \neq 1$ .

Suppose first that $b^u = b^v$ . Apply $\log_b$ to both sides. A function gives one output per input, so feeding it the same number twice gives the same answer twice, and $\log_b(b^u) = \log_b(b^v)$ . The first cancellation identity turns the left side into $u$ and the right side into $v$ , so $u = v$ . The converse is immediate: if $u = v$ , then $b^u$ and $b^v$ are the very same number. So $b^u = b^v$ holds exactly when $u = v$ , with no side conditions at all, since $b^t$ is defined for every real $t$ .

Notice where the requirement $b \neq 1$ earns its keep. Since $1^u = 1$ for every $u$ , we have $1^2 = 1^5$ while $2 \neq 5$ . A base of $1$ repeats a value, so it is not one-to-one and it has no undo button. The same goes for $b = 0$ , since $0^t = 0$ for every $t > 0$ . A negative base fails for a different reason, and $b = -2$ shows it plainly: $(-2)^{1/2}$ is not a real number at all, so $b^t$ is not even defined for every real $t$ , and there is no unbroken curve to invert in the first place. That is why logarithms are built only on bases that are positive and different from $1$ .

Now suppose $M$ and $N$ are positive and $\log_b M = \log_b N$ . Raise $b$ to each side: $b^{\log_b M} = b^{\log_b N}$ . The second cancellation identity turns the left side into $M$ and the right side into $N$ , so $M = N$ . Here the converse needs that positivity. If $M = N$ and both are positive, then $\log_b M$ and $\log_b N$ are the same number and the log equation holds. Drop the positivity and the converse collapses, because $\log_b(-4)$ is not a number at all, so $M = N = -4$ does not produce a true statement about logarithms.

That asymmetry is worth staring at. Equating exponents is a two-way street. Equating arguments is a two-way street only on positive arguments, and every extraneous solution in this lesson comes from stepping off the street.

The graph of y = 2 to the x climbs steadily and never repeats a value, so a horizontal line meets it at most once. That is the whole reason 2 to the u equals 2 to the v forces u to equal v.

Solving exponential equations

An exponential equation has the unknown in an exponent. There are two techniques, and which one you reach for depends on whether the two sides can be written with a common base.

Technique 1: match the bases. If both sides are powers of the same base, the theorem lets you throw the base away and set the exponents equal. For $8^x = 4^{x+1}$ , write everything in base $2$ :

2^{3x} = 2^{2x + 2} \;\Longrightarrow\; 3x = 2x + 2 \;\Longrightarrow\; x = 2.

Worked example 1 Solve $9^{x+1} = 27^{x-1}$

Neither side is a whole-number power of the other, but $9$ and $27$ are both powers of $3$ , so rewrite them:

9^{x+1} = \left(3^2\right)^{x+1} = 3^{2x+2}, \qquad 27^{x-1} = \left(3^3\right)^{x-1} = 3^{3x-3}.

Both sides now sit on the same base, so the exponents must be equal:

2x + 2 = 3x - 3 \;\Longrightarrow\; x = 5.

Check it in the original equation. With $x = 5$ , the left side is $9^6 = (3^2)^6 = 3^{12}$ and the right side is $27^4 = (3^3)^4 = 3^{12}$ , so both sides really are the same number.

Technique 2: take a logarithm of both sides. A common base is a luxury. In $5^x = 40$ there is none, because $40$ is not a whole-number power of $5$ that you can spot by inspection. Both sides are positive, though ( $5^x > 0$ for every $x$ , and $40 > 0$ ), so both sides have a logarithm. Applying the one-to-one function $\log$ to both sides therefore produces an equation with exactly the same solutions. Use the common logarithm $\log$ , which is $\log_{10}$ :

\log\!\left(5^x\right) = \log 40 \;\Longrightarrow\; x \log 5 = \log 40 \;\Longrightarrow\; x = \frac{\log 40}{\log 5}.

The middle step is the power rule from Properties of Logarithms. That power rule is the whole point of taking a log: it drags the unknown down out of the exponent, where ordinary algebra can reach it.

That quotient is the change-of-base formula in action. Setting $y = \log_5 40$ means $5^y = 40$ , and taking $\log$ of both sides gives $y \log 5 = \log 40$ , which is the computation we just did. So

\log_b M = \frac{\log M}{\log b},

and change of base is not a separate trick but the same move applied to the definition. Numerically, $\log 40 \approx 1.60206$ and $\log 5 \approx 0.69897$ , so

x = \frac{\log 40}{\log 5} \approx 2.29203.

The exact answer is $x = \log_5 40$ . Rounded to three decimal places, $x \approx 2.292$ . Keep those two statements apart: $\log_5 40$ is the exact value, and $2.292$ is a decimal approximation of it, close but not identical, which is exactly what $\approx$ signals instead of $=$ .

Worked example 2 Solve $4 \cdot 7^{x-1} = 300$

The power is buried under a coefficient, so isolate it first by dividing both sides by $4$ :

7^{x-1} = 75.

Now $75$ is not a whole-number power of $7$ , so match-the-bases is out. Take the common logarithm of both sides (legal, since both sides are positive) and use the power rule:

(x - 1)\log 7 = \log 75 \;\Longrightarrow\; x - 1 = \frac{\log 75}{\log 7} = \log_7 75.

So the exact solution is $x = 1 + \log_7 75$ . For a decimal, use $\log 75 \approx 1.87506$ and $\log 7 \approx 0.84510$ :

x \approx 1 + \frac{1.87506}{0.84510} \approx 1 + 2.21875 = 3.21875.

Rounded to three decimal places, $x \approx 3.219$ . As a sanity check, $7^{2.219} \approx 75$ , and multiplying by $4$ returns roughly $300$ .

Nothing here can create a false solution: every step (dividing by $4$ , taking a log of two positive quantities, dividing by $\log 7$ ) is reversible.

Check your understanding

Solve $6^x = 50$ . Which option gives both the exact solution and its correct three-decimal-place approximation?

x = \log_6 50 \approx 2.183

x = \log_6 50 \approx 8.333

x = \log_{50} 6 \approx 0.458

x = 50 - 6 \approx 44.000

Check your understanding

Solve $2^{3x} = 32$ .

x = \dfrac{5}{3}

x = 5

x = \dfrac{32}{3}

x = \dfrac{3}{5}

Solving logarithmic equations

A logarithmic equation has the unknown inside a logarithm. Again there are two shapes.

Shape A: one logarithm equal to a constant. Rewrite $\log_b u = k$ in exponential form, $u = b^k$ . This step is completely safe, and the reason is worth spelling out: $b^k$ is positive no matter what $k$ is. So any $x$ satisfying $u(x) = b^k$ automatically satisfies $u(x) > 0$ , which is exactly what the original equation needed. The candidate cannot fail the domain test. For example,

\log_2(3x - 1) = 4 \;\Longrightarrow\; 3x - 1 = 2^4 = 16 \;\Longrightarrow\; x = \frac{17}{3},

and indeed $3x - 1 = 16 > 0$ , as guaranteed. You may check anyway, but this particular check cannot fail.

Shape B: one logarithm equal to another. By the theorem, $\log_b M = \log_b N$ gives $M = N$ . But the converse only holds when $M$ is positive, so the equation $M = N$ can carry solutions the original never had. Here the check is not optional. Take

\log_3\!\left(x^2 - 6\right) = \log_3 x.

Equating the arguments gives $x^2 - 6 = x$ , so $x^2 - x - 6 = 0$ , so $(x - 3)(x + 2) = 0$ , and the candidates are $x = 3$ and $x = -2$ . Test them in the original equation. At $x = 3$ the arguments are $3^2 - 6 = 3$ and $3$ , both positive, and both sides equal $\log_3 3 = 1$ . At $x = -2$ the right side is $\log_3(-2)$ , which does not exist, so $x = -2$ is not a solution at all. The only solution is $x = 3$ .

Check your understanding

Solve $\log_5(2x + 3) = 2$ .

x = 1

x = \dfrac{7}{2}

x = 22

x = 11

Extraneous solutions, and where they come from

You met this mechanism in Solving Radical Equations. Squaring both sides is a one-way step: from $u = v$ you may conclude $u^2 = v^2$ , but $u^2 = v^2$ only gives back $u = \pm v$ . Squaring destroys sign information, so the squared equation can hold at points where the original fails. The extra roots are not mistakes in your algebra; the step itself manufactured them.

Logarithms do the same thing, with the domain playing the role that the sign played. The product rule $\log_b M + \log_b N = \log_b(MN)$ was proved for $M > 0$ and $N > 0$ . If both are negative, the left side is meaningless while the right side is perfectly happy, since a product of two negatives is positive. Rewriting the left side as the right side can therefore enlarge the set of $x$ where the equation makes sense, and that is exactly what happens below.

A combined equation can never lose a root: wherever the original sum of logs is defined, both arguments are positive, so the product rule turns one equation into the other and a genuine solution survives the rewrite. It can only ever add a root, and only at an $x$ where both arguments have become negative, since that is the one way the product can be positive while the sum was never defined at all.

Equating arguments has the same one-way character, for the same reason: $\log_b M = \log_b N$ gives $M = N$ , but $M = N$ hands back the log equation only when $M > 0$ . Candidates from that step need the same test, as $x = -2$ discovered a moment ago.

Exponential equations are the honest contrast. Taking a log of both sides of $5^x = 40$ is reversible because both sides there are already positive, and $\log$ is one-to-one, so nothing is gained or lost. Whenever an exponential expression is isolated on one side and a positive number sits on the other, that same reasoning applies and the step never produces an extraneous root. It is the logarithmic side of this lesson that needs the domain check.

For the equation log base 2 of x plus log base 2 of (x minus 2) equals 3, combining the two logarithms widens the set of x values where the equation makes sense. The arrows show that the shaded regions run on without end. The candidate x = -2 lives only in the region that the combining step added, so it is extraneous, while x = 4 lies in both regions and is a genuine solution.

Worked example 3 Solve $\log_2 x + \log_2(x - 2) = 3$

Write down the domain before doing any algebra. The equation asks for $\log_2 x$ and $\log_2(x-2)$ , so it needs $x > 0$ and $x - 2 > 0$ . Both conditions together say

x > 2.

Now combine the two logarithms with the product rule and convert to exponential form:

\log_2\!\big(x(x - 2)\big) = 3 \;\Longrightarrow\; x(x - 2) = 2^3 = 8.

That is a quadratic. Expand and factor:

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

so the candidates are $x = 4$ and $x = -2$ .

Test each one in the original equation. For $x = 4$ the arguments are $4 > 0$ and $4 - 2 = 2 > 0$ , and

\log_2 4 + \log_2 2 = 2 + 1 = 3,

so $x = 4$ is genuine. For $x = -2$ the first argument is $-2$ , and $\log_2(-2)$ does not exist, so $x = -2$ is rejected. It also fails the domain condition $x > 2$ that we wrote down at the start.

At $x = -2$ the two arguments are $x = -2$ and $x - 2 = -4$ , both negative, and their product is $(-2)(-4) = 8$ , precisely the value the combined equation wanted. The combined equation is satisfied there; the original one is not even defined there. The solution is $x = 4$ .

Check your understanding

What is the complete solution set of $\log_4 x + \log_4(x - 6) = 2$ ?

\{-2,\; 8\}

\{8\}

\{-2\}

D The equation has no solution.

Equations that hide a quadratic

Some exponential equations look unsolvable until you notice that one power is the square of another. Since $4^x = (2^2)^x = (2^x)^2$ , an equation containing both $4^x$ and $2^x$ is a quadratic wearing a disguise. Substitute $u = 2^x$ , solve the quadratic, then undo the substitution.

Worked example 4 Solve $4^x - 3 \cdot 2^x - 4 = 0$

Rewrite the first term so that both powers are built from $2^x$ :

4^x = \left(2^2\right)^x = 2^{2x} = \left(2^x\right)^2.

Let $u = 2^x$ . The equation becomes an ordinary quadratic in $u$ , which factors:

u^2 - 3u - 4 = 0 \;\Longrightarrow\; (u - 4)(u + 1) = 0 \;\Longrightarrow\; u = 4 \ \text{ or } \ u = -1.

Undo the substitution one root at a time. From $u = 4$ we get $2^x = 4 = 2^2$ , and equating exponents gives $x = 2$ .

From $u = -1$ we get $2^x = -1$ , which has no solution: an exponential is positive for every real input, so $2^x$ never reaches $-1$ . Notice this is a different situation from an extraneous root. A candidate $x$ was never produced at all, because $-1$ is outside the range of $2^x$ ; nothing needs to be rejected after the fact.

Check the survivor in the original equation:

4^2 - 3 \cdot 2^2 - 4 = 16 - 12 - 4 = 0.

The equation holds, so the solution is $x = 2$ .

Check your understanding

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

x = 2

x = 2

and

x = -1

x = -1

D No solution

Common mistakes

Skipping the domain test after combining logs. The combined equation is a consequence of the original, not an equivalent of it, so it can carry extra roots. Always test each candidate in the original equation, where every logarithm’s argument must be positive.
Rejecting negative candidates on sight. The test is whether each argument is positive, not whether $x$ is. In $\log_2(x^2 - 9) = 4$ the value $x = -5$ is a perfectly good solution, since $(-5)^2 - 9 = 16 > 0$ . Throwing it away because it is negative loses half the answer.
Inventing $\log_b(M + N) = \log_b M + \log_b N$ . The product rule turns a sum of logs into the log of a product. There is no rule at all for the log of a sum.
Splitting a quotient of logs. $\dfrac{\log 40}{\log 5}$ is not $\log\dfrac{40}{5} = \log 8$ . The first is about $2.292$ and the second is about $0.903$ . Change of base produces a quotient of two logarithms, and it stays that way.
Equating exponents across different bases. From $3^x = 5^{x-1}$ you may not conclude $x = x - 1$ . A shared base is the entire license for dropping it, so with different bases you must take a logarithm instead.
Pulling a coefficient out of a logarithm. Writing $\log(4 \cdot 7^{x-1}) = 4 \log 7^{x-1}$ is wrong. Only an exponent comes out in front; a factor becomes a separate logarithm, so the correct expansion is $\log 4 + (x - 1)\log 7$ . Taking a log before isolating the power is perfectly legal and reaches the same answer, but dividing by $4$ first keeps the coefficient out of the way.
Trying to solve $2^x = -8$ . An exponential is always positive, so there is no solution and $\log_2(-8)$ is not a number. Write “no solution” rather than a symbol that does not exist.

Key takeaways

For $b > 0$ with $b \neq 1$ , the exponential $b^x$ and the logarithm $\log_b$ are one-to-one. So $b^u = b^v$ exactly when $u = v$ , and, for positive $M$ and $N$ , $\log_b M = \log_b N$ exactly when $M = N$ . Those two facts license every move in this lesson.
Exponential equation with a common base: rewrite both sides on that base and equate the exponents.
Exponential equation without one: isolate the exponential expression, take a logarithm of both sides, and use the power rule to bring the unknown down, then $\log_b M = \frac{\log M}{\log b}$ for a decimal. The exponential side is automatically positive; once the other side is positive too, this step is reversible and no extraneous roots appear.
A single log equal to a constant: $\log_b u = k$ becomes $u = b^k$ . Since $b^k > 0$ , the candidate is automatically inside the domain, so this step cannot manufacture a root either.
Combining logs, or equating arguments, can enlarge the domain, so either move can manufacture roots. Solve the easy combined equation, then test every candidate in the original equation: each argument must be positive. This is the same one-way-step trap as squaring in a radical equation.
Since $b^x > 0$ for every $x$ , a substitution value $u \leq 0$ yields no $x$ at all. That is a root of the quadratic that produces no solution, which is not the same as an extraneous solution.
State rounded answers as rounded. The exact solution of $5^x = 40$ is $\log_5 40$ ; the number $2.292$ is that solution to three decimal places.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Logarithms (Khan Academy)
Exponential and logarithmic equations (free textbook) (OpenStax)
Solving exponential equations (Paul's Online Math Notes)
Solving logarithm equations (Paul's Online Math Notes)
Working with exponents and logarithms (Math is Fun)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

The general proof that combining logarithms can add roots but never lose them

Combining logarithms can add roots, but never lose them#

Let $M(x)$ and $N(x)$ be expressions in $x$ , and compare the equation you are given,

\log_b M(x) + \log_b N(x) = k,

with the equation you get after combining,

\log_b\!\big(M(x)\,N(x)\big) = k.

The first equation makes sense only where $M(x) > 0$ and $N(x) > 0$ . The second makes sense wherever $M(x)N(x) > 0$ , which happens when both factors are positive and also when both factors are negative. So the second equation’s domain contains the first equation’s domain, and it is strictly larger as soon as some $x$ makes both factors negative.

Take any solution of the first equation. There $M > 0$ and $N > 0$ , so the product rule applies and the left sides of the two equations are equal at that $x$ ; hence that $x$ solves the second equation too. Combining cannot lose a root.

The reverse implication can only fail on the region where the domains differ. If $x$ solves the second equation with $M(x) < 0$ and $N(x) < 0$ , then the first equation is not even defined there, so $x$ is not one of its solutions. Such an $x$ is an extraneous root, produced by the step rather than by the problem. (Most of the added region causes no trouble at all, because nothing there solves the second equation either. The damage is done only by an added point that happens to be a root.)

That gives you a rule you can trust. Combine and solve, because the combined equation is the easy one. Then test every candidate in the original equation, where the test is simply whether each logarithm’s argument is positive. Keep the survivors and discard the rest.

A bit of history (optional)

This lesson keeps telling you to throw an answer away, and the reason is always the same. A logarithm has been handed a negative number, and there is no such thing. Two of the best minds of their generation once spent a year arguing about that.

The quarrel ran by letter, through 1712 and 1713, between Gottfried Leibniz and Johann Bernoulli. Bernoulli had a proof that looked airtight, and it used the power rule you already know. Since $(-1)^{2} = 1^{2}$ , taking logarithms of both sides gives $2\log(-1) = 2\log 1$ . So $\log(-1)$ has to be $\log 1$ , which is $0$ .

Leibniz refused it: a logarithm is an exponent, he wrote back. No exponent on a positive base ever gives a negative number, so there is nothing there to be $0$ . Neither man gave way, and the quarrel outlived them both.

It was settled a generation later, and Leibniz was right about the real numbers. Bernoulli’s step was the illegal one, because the power rule requires a positive argument. Applied to a negative one, it manufactures a value out of nothing.

That is the mechanism behind an extraneous root, in miniature. Folding two logarithms into one quietly widens the set of numbers your equation will accept. A candidate that survives the algebra may be standing on ground the original equation never covered. So put it back in and check what each logarithm is holding. Bernoulli did not.