The Central Idea

Exponential and logarithmic equations are solved by undoing their defining operations.

\[ b^x=y \quad\Longleftrightarrow\quad \boxed{x=\log_b(y)} \]

and

\[ \log_b(x)=y \quad\Longleftrightarrow\quad \boxed{x=b^y}. \]

These equations express the same inverse relationship in opposite directions.

Identify the Equation Type First

Recognizing the structure often determines the cleanest solution method.

Method 1: Match Exponential Bases

For a valid exponential base,

\[ b^u=b^v \]

implies

\[ \boxed{u=v}. \]

This follows from the one-to-one property of exponential functions.

Example: Matching Bases

Solve

\[ 2^{x+1}=16. \]

Since

\[ 16=2^4, \]

we have

\[ 2^{x+1}=2^4. \]

Therefore,

\[ x+1=4 \]

and

\[ \boxed{x=3}. \]

Example: Matching a Reciprocal Power

Solve

\[ 4^x=\frac1{64}. \]

Rewrite with base \(2\):

\[ (2^2)^x=2^{-6}. \]

Therefore,

\[ 2^{2x}=2^{-6} \]

and

\[ 2x=-6. \]

Thus,

\[ \boxed{x=-3}. \]

Example: Matching a Radical

Solve

\[ 9^x=\sqrt3. \]

Rewrite:

\[ (3^2)^x=3^{1/2}. \]

Therefore,

\[ 2x=\frac12. \]

Hence,

\[ \boxed{x=\frac14}. \]

When Matching Bases Is Not Convenient

Consider

\[ 3^x=20. \]

The value \(20\) is not a convenient integer power of \(3\). This is the natural point to use logarithms.

Method 2: Take Logarithms of Both Sides

From

\[ b^x=y, \]

take natural logarithms:

\[ \ln(b^x)=\ln y. \]

Apply the logarithm power law:

\[ x\ln b=\ln y. \]

Therefore,

\[ \boxed{x=\frac{\ln y}{\ln b}}. \]

Example: Solve \(3^x=20\)

\[ \ln(3^x)=\ln20. \]

Therefore,

\[ x\ln3=\ln20 \]

and

\[ x=\frac{\ln20}{\ln3}. \]

Thus,

\[ \boxed{x\approx2.7268}. \]

Common Logarithms Work Too

The same equation can be solved using base-\(10\) logarithms:

\[ \log(3^x)=\log20. \]

Therefore,

\[ x=\frac{\log20}{\log3}. \]

Change of base guarantees the same result.

Isolate the Exponential Expression First

In

\[ 5(2^x)=40, \]

divide by \(5\) first:

\[ 2^x=8. \]

Therefore,

\[ \boxed{x=3}. \]

Taking logarithms before isolating the exponential usually creates unnecessary work.

Example: Isolate before Logging

Solve

\[ 7(1.12)^t=49. \]

Divide by \(7\):

\[ (1.12)^t=7. \]

Take logarithms:

\[ t\ln(1.12)=\ln7. \]

Therefore,

\[ \boxed{ t=\frac{\ln7}{\ln(1.12)} \approx17.1705 }. \]

Example: Unknown Time in a Growth Model

Suppose

\[ P(t)=1200(1.08)^t. \]

Find the time when \(P(t)=5000\):

\[ 5000=1200(1.08)^t. \]

Divide:

\[ \frac{5000}{1200}=(1.08)^t. \]

Therefore,

\[ t= \frac{\ln(5000/1200)}{\ln(1.08)}. \]

Thus,

\[ \boxed{t\approx18.5434}. \]

Example: Unknown Time in a Decay Model

A quantity retains \(87\%\) of its value each year. Find the time required to reach half its initial value.

\[ (0.87)^t=\frac12. \]

Therefore,

\[ t= \frac{\ln(1/2)}{\ln(0.87)}. \]

Thus,

\[ \boxed{t\approx4.9773}. \]

Equations with Base \(e\)

Natural logarithms are especially convenient for powers of \(e\):

\[ \boxed{\ln(e^x)=x}. \]

This makes \(\ln\) the natural inverse operation for continuous exponential models.

Example: Solve an Equation with \(e\)

Solve

\[ e^{0.18t}=9. \]

Take natural logarithms:

\[ 0.18t=\ln9. \]

Therefore,

\[ \boxed{ t=\frac{\ln9}{0.18}\approx12.2068 }. \]

Example: Time Required to Triple

Suppose

\[ A(t)=A_0e^{0.07t}. \]

To triple,

\[ 3A_0=A_0e^{0.07t}. \]

Divide by \(A_0\):

\[ 3=e^{0.07t}. \]

Therefore,

\[ \boxed{ t=\frac{\ln3}{0.07}\approx15.6945 }. \]

Expressions Can Appear in the Exponent

Solve

\[ 2^{3x-1}=17. \]

Take logarithms:

\[ (3x-1)\ln2=\ln17. \]

Therefore,

\[ 3x-1=\frac{\ln17}{\ln2}. \]

Solve the resulting linear equation:

\[ \boxed{ x= \frac{1+\ln17/\ln2}{3} }. \]

Some Exponential Equations Become Quadratics

Consider

\[ 4^x-5(2^x)+4=0. \]

Since

\[ 4^x=(2^x)^2, \]

let

\[ u=2^x. \]

Then

\[ u^2-5u+4=0. \]

Example: Exponential Substitution

Factor:

\[ (u-1)(u-4)=0. \]

Thus

\[ u=1 \]

or

\[ u=4. \]

Substitute back:

\[ 2^x=1\Rightarrow x=0 \]

and

\[ 2^x=4\Rightarrow x=2. \]

Therefore,

\[ \boxed{x=0,\;2}. \]

Now Turn to Logarithmic Equations

A logarithmic equation contains a variable inside a logarithm.

The basic inverse relationship is

\[ \boxed{ \log_b(M)=c \Longleftrightarrow M=b^c }. \]

Method 3: Convert One Logarithm to Exponential Form

Solve

\[ \log_3(x)=4. \]

Convert:

\[ x=3^4. \]

Therefore,

\[ \boxed{x=81}. \]

Example: Variable Expression inside a Logarithm

Solve

\[ \log_2(x-3)=5. \]

Convert:

\[ x-3=2^5=32. \]

Therefore,

\[ \boxed{x=35}. \]

Logarithmic Domain Restrictions Come First

The expression

\[ \log_2(x-3) \]

requires

\[ x-3>0. \]

Therefore,

\[ \boxed{x>3}. \]

The candidate \(x=35\) is valid.

Method 4: Equal Logarithms Have Equal Arguments

For the same valid logarithmic base,

\[ \log_b(M)=\log_b(N) \]

implies

\[ \boxed{M=N}, \]

provided both arguments are positive.

Example: Equal Logarithms

Solve

\[ \log_5(x+4)=\log_5(3x-8). \]

Equate arguments:

\[ x+4=3x-8. \]

Therefore,

\[ x=6. \]

Check:

\[ x+4=10>0,\qquad3x-8=10>0. \]

Thus,

\[ \boxed{x=6}. \]

Method 5: Condense Multiple Logarithms

If several logarithms with the same base appear on one side, logarithm laws can combine them before conversion:

\[ \log_b(M)+\log_b(N)=\log_b(MN) \]

and

\[ \log_b(M)-\log_b(N) = \log_b\left(\frac{M}{N}\right). \]

Example: Sum of Logarithms

Solve

\[ \log_2(x)+\log_2(x-2)=3. \]

Condense:

\[ \log_2(x(x-2))=3. \]

Convert:

\[ x(x-2)=8. \]

Therefore,

\[ x^2-2x-8=0. \]

Reject the Extraneous Candidate

Factor:

\[ (x-4)(x+2)=0. \]

Candidates are

\[ x=4,\qquad x=-2. \]

The original logarithms require

\[ x>2. \]

Therefore only

\[ \boxed{x=4} \]

is valid.

Extraneous Solutions Are Especially Important with Logarithms

Algebraic manipulation may produce candidates that make an original logarithmic argument zero or negative.

Those candidates are not solutions to the original equation and must be rejected.

Example: Difference of Logarithms

Solve

\[ \log_3(x)-\log_3(x-2)=1. \]

Condense:

\[ \log_3\left(\frac{x}{x-2}\right)=1. \]

Convert:

\[ \frac{x}{x-2}=3. \]

Therefore,

\[ x=3x-6 \]

and

\[ \boxed{x=3}. \]

Example: A Coefficient on a Logarithm

Solve

\[ 2\log_5(x)=4. \]

Divide by \(2\):

\[ \log_5(x)=2. \]

Convert:

\[ x=25. \]

Therefore,

\[ \boxed{x=25}. \]

Power-Law Route and Domain Check

Starting again with

\[ 2\log_5(x)=4, \]

the power law gives

\[ \log_5(x^2)=4. \]

Therefore,

\[ x^2=625. \]

Algebra gives \(x=\pm25\), but the original logarithm requires

\[ x>0. \]

Thus only

\[ \boxed{x=25} \]

survives the domain check.

Natural Logarithm Equations

Since

\[ \ln x=\log_e(x), \]

the equation

\[ \ln x=2 \]

becomes

\[ x=e^2. \]

Therefore,

\[ \boxed{x=e^2}. \]

Example: Shifted Natural Logarithm

Solve

\[ \ln(x+1)=3. \]

Exponentiate:

\[ x+1=e^3. \]

Therefore,

\[ \boxed{x=e^3-1}. \]

Exact Form versus Approximate Form

The solution

\[ x=\frac{\ln20}{\ln3} \]

is exact.

The decimal

\[ x\approx2.7268 \]

is approximate.

Keep exact logarithmic forms until a decimal is needed.

Round at the End

Logarithmic calculations often contain irrational intermediate values. Rounding those values early can accumulate error.

Enter the full expression whenever practical and round only the final result.

Checking an Exponential Solution

If \(x=3\) is proposed for

\[ 2^{x+1}=16, \]

substitute:

\[ 2^{3+1}=2^4=16. \]

The solution checks.

Checking a Logarithmic Solution

If \(x=4\) is proposed for

\[ \log_2(x)+\log_2(x-2)=3, \]

then

\[ \log_2(4)+\log_2(2)=2+1=3. \]

Both arguments are positive, so the solution is valid.

Graphical Meaning of an Exponential Equation

Solving

\[ 3^x=20 \]

asks where the graph

\[ y=3^x \]

intersects the horizontal line

\[ y=20. \]

The logarithmic solution

\[ x=\log_3(20) \]

gives that intersection coordinate exactly.

Graphical Meaning of a Logarithmic Equation

Solving

\[ \log_2(x)=5 \]

asks where

\[ y=\log_2(x) \]

intersects \(y=5\).

The intersection occurs at

\[ \boxed{x=32}. \]

Some Exponential Equations Have No Real Solution

Consider

\[ 2^x=-3. \]

Since

\[ 2^x>0 \]

for every real \(x\), the equation has

\[ \boxed{\text{no real solution}}. \]

Some Logarithmic Equations Have No Solution

Consider

\[ \log_2(x-1)=\log_2(x-3). \]

Equal logarithms would require

\[ x-1=x-3, \]

which is impossible.

Therefore,

\[ \boxed{\text{no solution}}. \]

Common Error: Taking Logarithms Too Early

In

\[ 4+3(2^x)=28, \]

first isolate the exponential:

\[ 3(2^x)=24 \]

and then

\[ 2^x=8. \]

Simplifying first is usually cleaner.

Common Error: Pulling an Exponent Down without a Logarithm

From

\[ b^x=y, \]

it is incorrect to replace the left side with \(bx\).

The exponent comes down only after applying a logarithm:

\[ \log(b^x)=x\log b. \]

Common Error: Applying a Logarithm to Only One Side

From

\[ 3^x=20, \]

write

\[ \ln(3^x)=\ln20. \]

The same operation must be applied to both sides.

Common Error: Splitting a Logarithm of a Sum

In

\[ \log(x+5)=2, \]

do not replace the left side with

\[ \log x+\log5. \]

Logarithm laws do not split addition.

Common Error: Forgetting the Domain Check

Any candidate that makes an original logarithmic argument zero or negative must be rejected.

This is especially important after factoring, using the quadratic formula, or applying the logarithm power law.

Common Error: Confusing Exact and Approximate Equality

If

\[ x=\frac{\ln20}{\ln3}, \]

then the rounded decimal should be written

\[ \boxed{x\approx2.7268}. \]

A Reliable Procedure for Exponential Equations

  1. Simplify and isolate the exponential expression.
  2. Check whether both sides can be written with the same base.
  3. If so, equate exponents.
  4. If not, take logarithms of both sides.
  5. Use the power law to bring the exponent down.
  6. Solve the resulting algebraic equation.
  7. Keep an exact form until the final approximation.
  8. Check the result when useful.

A Reliable Procedure for Logarithmic Equations

  1. Record the positivity restriction for every logarithmic argument.
  2. If there is one logarithm, convert directly to exponential form.
  3. If equal logarithms share a base, equate their arguments.
  4. If several logarithms occur on one side, condense them first.
  5. Convert the resulting logarithmic equation to exponential form.
  6. Solve the resulting algebraic equation.
  7. Reject candidates outside the original logarithmic domain.
  8. Check the remaining candidates in the original equation.

Example: Full Exponential Workflow

Solve

\[ 200(1.03)^t=900. \]

Divide:

\[ (1.03)^t=4.5. \]

Take logarithms:

\[ t\ln(1.03)=\ln4.5. \]

Therefore,

\[ t= \frac{\ln4.5}{\ln1.03}. \]

Thus,

\[ \boxed{t\approx50.8842}. \]

Example: Full Logarithmic Workflow

Solve

\[ \log_3(x-1)+\log_3(x-3)=2. \]

The domain requires

\[ x>3. \]

Condense:

\[ \log_3((x-1)(x-3))=2. \]

Convert:

\[ (x-1)(x-3)=9. \]

Rearrange:

\[ x^2-4x-6=0. \]

The quadratic formula gives

\[ x=2\pm\sqrt{10}. \]

Since \(x>3\), only

\[ \boxed{x=2+\sqrt{10}} \]

is valid.

Practice 1: Matching Bases

Solve

\[ 3^{x+2}=243. \]

Practice 2: Reciprocal Power

Solve

\[ 5^{2x}=\frac1{125}. \]

Practice 3: Use Logarithms

Solve

\[ 7^x=30. \]

Give an exact form and a decimal approximation.

Practice 4: Isolate an Exponential

Solve

\[ 4(2^x)=60. \]

Practice 5: Equation with \(e\)

Solve

\[ e^{0.25t}=6. \]

Practice 6: Basic Logarithmic Equation

Solve

\[ \log_4(x)=3. \]

Practice 7: Shifted Logarithmic Equation

Solve

\[ \log_2(x+5)=4. \]

Practice 8: Equal Logarithms

Solve

\[ \ln(x+2)=\ln(3x-4). \]

Practice 9: Condense and Solve

Solve

\[ \log_2(x)+\log_2(x-2)=4. \]

Practice 10: Domain Check

Solve

\[ \log_3(x-1)=\log_3(5-x). \]

Verify the result against both logarithmic arguments.

Practice Answers

Practice 1

Since \(243=3^5\),

\[ x+2=5. \]

Therefore,

\[ \boxed{x=3}. \]

Practice 2

Since

\[ \frac1{125}=5^{-3}, \]

we have

\[ 2x=-3. \]

Therefore,

\[ \boxed{x=-\frac32}. \]

Practice 3

\[ x=\frac{\ln30}{\ln7}. \]

Therefore,

\[ \boxed{x\approx1.7479}. \]

Practice 4

\[ 2^x=15. \]

Therefore,

\[ x=\frac{\ln15}{\ln2} \]

and

\[ \boxed{x\approx3.9069}. \]

Practice 5

\[ 0.25t=\ln6. \]

Therefore,

\[ \boxed{t=4\ln6\approx7.1670}. \]

Practice 6

\[ x=4^3. \]

Therefore,

\[ \boxed{x=64}. \]

Practice 7

\[ x+5=16. \]

Therefore,

\[ \boxed{x=11}. \]

Practice 8

\[ x+2=3x-4. \]

Therefore,

\[ \boxed{x=3}. \]

Both original arguments equal \(5\), so the solution is valid.

Practice 9

Condense:

\[ \log_2(x(x-2))=4. \]

Convert:

\[ x(x-2)=16. \]

Therefore,

\[ x^2-2x-16=0. \]

The candidates are

\[ x=1\pm\sqrt{17}. \]

The original equation requires \(x>2\), so only

\[ \boxed{x=1+\sqrt{17}} \]

is valid.

Practice 10

\[ x-1=5-x. \]

Therefore,

\[ x=3. \]

Check:

\[ x-1=2>0,\qquad5-x=2>0. \]

Thus,

\[ \boxed{x=3}. \]

What to Remember

Exponential equations are solved by exposing or undoing the exponent. Logarithmic equations are solved by undoing the logarithm or condensing logarithms into a form that can be exponentiated.

The central habit is: identify the outermost operation involving the unknown, undo that structure carefully, and preserve the original domain throughout the solution.

Where We Go Next

Exponential equations make clear that multiplicative change behaves very differently from additive change.

The next lesson, Linear versus Exponential Change, compares these two major patterns through formulas, tables, graphs, differences, ratios, and long-run behavior so that each model can be recognized from the structure of a problem.