The Central Idea
Consider
\[ 2^x=8. \]Since
\[ 8=2^3, \]we can rewrite the equation as
\[ 2^x=2^3. \]Because the exponential function with base \(2\) is one-to-one, equal outputs require equal exponents:
\[ x=3. \]But not every exponential equation can be rewritten with a convenient common base.
For example,
\[ 2^x=7 \]has no simple integer exponent. In that case, logarithms let us isolate the exponent:
\[ x=\frac{\ln 7}{\ln 2}. \]The method depends on the structure of the equation.
Exponential Equations
An exponential equation contains a variable in an exponent.
Examples include
\[ 3^x=81, \] \[ 5^{2x-1}=17, \]and
\[ 2^x+2^{x+1}=24. \]The first question is usually:
Can both sides be rewritten with the same base?
The One-to-One Property of Exponential Functions
If
\[ a>0,\qquad a\ne1, \]then the function
\[ f(x)=a^x \]is one-to-one.
Therefore,
\[ a^u=a^v \quad\Longleftrightarrow\quad u=v. \]This property is the foundation of the common-base method.
Common-Base Method
Solve
\[ 3^x=27. \]Rewrite \(27\):
\[ 27=3^3. \]Then
\[ 3^x=3^3. \]Therefore,
\[ x=3. \]Rewrite Both Sides
Solve
\[ 8^{x+1}=4^{2x}. \]Rewrite both bases as powers of \(2\):
\[ 8=2^3, \qquad 4=2^2. \]Then
\[ (2^3)^{x+1}=(2^2)^{2x}. \]Use the power-of-a-power rule:
\[ 2^{3x+3}=2^{4x}. \]Equate exponents:
\[ 3x+3=4x. \]So
\[ x=3. \]Negative Exponents and Reciprocals
Solve
\[ 5^{-x}=\frac{1}{125}. \]Since
\[ 125=5^3, \]we have
\[ \frac{1}{125}=5^{-3}. \]Therefore,
\[ 5^{-x}=5^{-3}, \]so
\[ -x=-3. \]Thus,
\[ x=3. \]Fractional Bases
Solve
\[ \left(\frac{1}{2}\right)^x=8. \]Rewrite
\[ \frac12=2^{-1} \]and
\[ 8=2^3. \]Then
\[ 2^{-x}=2^3. \]Therefore,
\[ -x=3, \]so
\[ x=-3. \]Isolate the Exponential Expression First
Consider
\[ 3\cdot2^x+5=29. \]Subtract \(5\):
\[ 3\cdot2^x=24. \]Divide by \(3\):
\[ 2^x=8. \]Rewrite \(8=2^3\):
\[ x=3. \]Just as with absolute-value and radical equations, isolate the special structure before applying its inverse operation.
When a Common Base Is Not Convenient
Solve
\[ 2^x=7. \]There is no convenient power of \(2\) equal to \(7\).
Take logarithms of both sides:
\[ \ln(2^x)=\ln7. \]Use the power rule:
\[ x\ln2=\ln7. \]Divide by \(\ln2\):
\[ x=\frac{\ln7}{\ln2}. \]This is an exact answer.
Why Taking Logarithms Works
The natural logarithm is one-to-one on
\[ (0,\infty). \]Exponential expressions with positive bases are positive, so if
\[ A=B \]with
\[ A>0,\qquad B>0, \]then
\[ \ln A=\ln B \]is an equivalent transformation.
Logarithms turn exponents into multiplicative factors, which makes the variable easier to isolate.
Logarithms with Any Valid Base
We do not have to use natural logarithms.
From
\[ 2^x=7, \]we could take base-\(10\) logarithms:
\[ \log(2^x)=\log7. \]Then
\[ x\log2=\log7, \]so
\[ x=\frac{\log7}{\log2}. \]Both forms are equal by the change-of-base relationship.
General Exponential Equation
For
\[ a^x=b, \]with
\[ a>0,\qquad a\ne1,\qquad b>0, \]taking logarithms gives
\[ x\ln a=\ln b. \]Therefore,
\[ x=\frac{\ln b}{\ln a}. \]This is the logarithmic definition of the base-\(a\) logarithm:
\[ x=\log_a b. \]Variable Expressions in the Exponent
Solve
\[ 5^{2x-1}=13. \]Take natural logarithms:
\[ \ln(5^{2x-1})=\ln13. \]Apply the power rule:
\[ (2x-1)\ln5=\ln13. \]Divide by \(\ln5\):
\[ 2x-1=\frac{\ln13}{\ln5}. \]Add \(1\):
\[ 2x=1+\frac{\ln13}{\ln5}. \]Therefore,
\[ x= \frac12 \left( 1+\frac{\ln13}{\ln5} \right). \]Exponential Equations with Repeated Structure
Consider
\[ 2^{2x}-5\cdot2^x+4=0. \]Notice that
\[ 2^{2x}=(2^x)^2. \]Let
\[ u=2^x. \]Then
\[ u^2-5u+4=0. \]Factor:
\[ (u-1)(u-4)=0. \]So
\[ u=1 \]or
\[ u=4. \]Return to \(2^x\):
\[ 2^x=1 \]or
\[ 2^x=4. \]Thus,
\[ x=0 \]or
\[ x=2. \]Factor Repeated Exponential Terms
Solve
\[ 3^x+3^{x+1}=36. \]Use
\[ 3^{x+1}=3\cdot3^x. \]Then
\[ 3^x+3\cdot3^x=36. \]Factor:
\[ 4\cdot3^x=36. \]So
\[ 3^x=9. \]Therefore,
\[ x=2. \]Logarithmic Equations
A logarithmic equation contains a variable inside a logarithm.
Examples include
\[ \log_2 x=5, \] \[ \ln(x-3)=2, \]and
\[ \log(x-1)+\log(x+2)=1. \]The first task is always to identify the logarithmic domain.
Logarithmic Domain Restriction
For real logarithms, the argument must be positive.
Thus,
\[ \log_a(x-4) \]requires
\[ x-4>0. \]Therefore,
\[ x>4. \]Every logarithmic argument in an equation must satisfy its own positivity condition.
Definition of a Logarithm
The relationship
\[ \log_a b=c \]means
\[ a^c=b, \]where
\[ a>0,\qquad a\ne1,\qquad b>0. \]This equivalence lets us convert logarithmic equations into exponential equations.
Basic Logarithmic Equation
Solve
\[ \log_2 x=5. \]Rewrite exponentially:
\[ 2^5=x. \]Therefore,
\[ x=32. \]The value is positive, so it satisfies the logarithmic domain.
Natural Logarithm Equation
Solve
\[ \ln(x-3)=2. \]Domain:
\[ x>3. \]Rewrite exponentially:
\[ x-3=e^2. \]Thus,
\[ x=e^2+3. \]This value satisfies the original domain restriction.
Equal Logarithms with the Same Base
Because a logarithmic function is one-to-one,
\[ \log_a M=\log_a N \]implies
\[ M=N, \]provided
\[ M>0,\qquad N>0. \]For example, solve
\[ \log_3(x+4)=\log_3(2x-1). \]Domain:
\[ x+4>0 \]and
\[ 2x-1>0. \]The second condition gives
\[ x>\frac12. \]Set the arguments equal:
\[ x+4=2x-1. \]Therefore,
\[ x=5. \]This satisfies the domain.
Product Rule for Logarithms
For positive \(M\) and \(N\),
\[ \log_a(MN) = \log_a M+\log_a N. \]This identity lets separate logarithms be combined before solving.
But the positivity requirements on the original logarithms must still be preserved.
Sum of Logarithms
Solve
\[ \log(x-1)+\log(x+2)=1, \]where \(\log\) denotes base \(10\).
Domain:
\[ x-1>0 \]and
\[ x+2>0. \]So
\[ x>1. \]Combine the logarithms:
\[ \log[(x-1)(x+2)]=1. \]Rewrite exponentially:
\[ (x-1)(x+2)=10. \]Expand:
\[ x^2+x-2=10. \]Then
\[ x^2+x-12=0. \]Factor:
\[ (x+4)(x-3)=0. \]Candidates:
\[ x=-4,\quad x=3. \]Only \(x=3\) satisfies \(x>1\).
Therefore,
\[ \boxed{x=3}. \]Quotient Rule for Logarithms
For positive \(M\) and \(N\),
\[ \log_a\left(\frac{M}{N}\right) = \log_a M-\log_a N. \]This can turn a difference of logarithms into one logarithm.
Difference of Logarithms
Solve
\[ \ln(x+3)-\ln x=\ln2. \]Domain:
\[ x+3>0 \]and
\[ x>0. \]So
\[ x>0. \]Combine:
\[ \ln\left(\frac{x+3}{x}\right)=\ln2. \]Since \(\ln\) is one-to-one,
\[ \frac{x+3}{x}=2. \]Then
\[ x+3=2x, \]so
\[ x=3. \]The value satisfies the domain.
Power Rule for Logarithms
For \(M>0\),
\[ \log_a(M^r)=r\log_a M. \]This is the rule that allows logarithms to bring exponents down as coefficients.
It is the main bridge from exponential equations to ordinary algebra.
Do Not Split Logarithms Across Addition
A common error is
\[ \log(a+b)=\log a+\log b. \]This is false in general.
Logarithm laws apply to products, quotients, and powers—not to sums and differences inside the argument.
Do Not Combine Logarithms Without Domain Conditions
The identity
\[ \ln A+\ln B=\ln(AB) \]requires
\[ A>0 \]and
\[ B>0. \]Even if the product \(AB\) is positive, the separate logarithms are not defined unless each original argument is positive.
The original domain remains authoritative.
Extraneous Solutions in Logarithmic Equations
Logarithmic equations often produce algebraic candidates outside the original domain.
Consider
\[ \log(x-2)+\log(x-5)=\log4. \]The domain requires
\[ x>5. \]Combining the logs gives
\[ \log[(x-2)(x-5)]=\log4. \]Thus,
\[ (x-2)(x-5)=4. \]Expanding:
\[ x^2-7x+10=4. \]So
\[ x^2-7x+6=0. \]Factor:
\[ (x-1)(x-6)=0. \]Candidates:
\[ x=1,\quad x=6. \]Only
\[ x=6 \]satisfies the original domain.
Exponential Equations Have Positive Outputs
If
\[ a>0, \]then
\[ a^x>0 \]for every real \(x\).
Therefore, an equation such as
\[ 2^x=-5 \]has no real solution.
Recognizing the range of an exponential function can classify some equations immediately.
Special Case: Base 1
The function
\[ 1^x \]is always \(1\).
So
\[ 1^x=1 \]is true for every real \(x\), while
\[ 1^x=3 \]has no solution.
This is why ordinary exponential-function definitions require
\[ a\ne1. \]Special Case: Exponential Equality with Different Bases
Consider
\[ 2^x=3^x. \]Divide by \(3^x\), which is always positive:
\[ \left(\frac{2}{3}\right)^x=1. \]For a positive base other than \(1\), the only real exponent producing \(1\) is
\[ x=0. \]Therefore,
\[ \boxed{x=0}. \]Exponential Equations from Growth Models
Suppose a quantity follows
\[ P=P_0e^{kt}. \]To solve for \(t\), divide by \(P_0\), assuming \(P_0\ne0\):
\[ \frac{P}{P_0}=e^{kt}. \]Take natural logarithms:
\[ \ln\left(\frac{P}{P_0}\right)=kt. \]Then
\[ t= \frac{1}{k} \ln\left(\frac{P}{P_0}\right), \qquad k\ne0. \]The logarithm converts exponential growth or decay into a linear equation in time.
Doubling Time
If
\[ P=P_0e^{kt} \]and we want the time when
\[ P=2P_0, \]then
\[ 2P_0=P_0e^{kt}. \]Assuming \(P_0\ne0\), divide by \(P_0\):
\[ 2=e^{kt}. \]Take natural logarithms:
\[ \ln2=kt. \]So the doubling time is
\[ t=\frac{\ln2}{k}. \]Half-Life
For exponential decay
\[ A=A_0e^{-kt}, \qquad k>0, \]the half-life satisfies
\[ \frac{A_0}{2}=A_0e^{-kt}. \]Divide by \(A_0\):
\[ \frac12=e^{-kt}. \]Take natural logarithms:
\[ \ln\left(\frac12\right)=-kt. \]Since
\[ \ln\left(\frac12\right)=-\ln2, \]we obtain
\[ t=\frac{\ln2}{k}. \]The same logarithmic structure appears in both doubling and halving problems.
Graphical Interpretation
Solving
\[ 2^x=7 \]means finding where the graphs
\[ y=2^x \]and
\[ y=7 \]intersect.
Because \(2^x\) is strictly increasing, there is exactly one real intersection.
Likewise, solving
\[ \ln x=2 \]means finding where
\[ y=\ln x \]intersects
\[ y=2. \]The inverse relationship between exponentials and logarithms explains why these equations undo one another.
Not Every Exponential Equation Has an Elementary Closed Form
Consider
\[ x+2^x=5. \]The variable appears both outside and inside an exponent.
Ordinary logarithm rules do not isolate \(x\) directly.
Such equations may require:
- graphical methods;
- numerical approximation;
- iteration;
- or, in some cases, special functions.
Recognizing when elementary algebra has reached its natural limit is part of equation solving.
Worked Example: Common Base
Solve
\[ 4^{x-1}=8. \]Rewrite in base \(2\):
\[ (2^2)^{x-1}=2^3. \]Then
\[ 2^{2x-2}=2^3. \]So
\[ 2x-2=3. \]Therefore,
\[ x=\frac52. \]Worked Example: Logarithmic Method
Solve
\[ 7^{3x}=20. \]Take natural logarithms:
\[ \ln(7^{3x})=\ln20. \]Then
\[ 3x\ln7=\ln20. \]Therefore,
\[ x=\frac{\ln20}{3\ln7}. \]Worked Example: Basic Logarithmic Equation
Solve
\[ \log_5(x+1)=3. \]Domain:
\[ x>-1. \]Rewrite exponentially:
\[ x+1=5^3. \]Thus,
\[ x+1=125, \]so
\[ x=124. \]The value satisfies the domain.
Worked Example: Two Logarithms
Solve
\[ \ln x+\ln(x-2)=\ln3. \]Domain:
\[ x>2. \]Combine:
\[ \ln[x(x-2)]=\ln3. \]Therefore,
\[ x(x-2)=3. \]Expand:
\[ x^2-2x-3=0. \]Factor:
\[ (x-3)(x+1)=0. \]Candidates:
\[ x=3,\quad x=-1. \]Only \(x=3\) lies in the domain.
Therefore,
\[ \boxed{x=3}. \]A Reliable Exponential Workflow
- Isolate the exponential expression when possible.
- Look for a common base.
- If a common base exists, rewrite and equate exponents.
- If not, take logarithms of both sides.
- Use logarithm laws to expose the exponent.
- Solve the resulting algebraic equation.
- Check the result when the original structure includes additional restrictions.
A Reliable Logarithmic Workflow
- Record the domain: every logarithmic argument must be positive.
- Simplify or combine logarithms when useful.
- Use one-to-one logarithm properties or rewrite in exponential form.
- Solve the resulting algebraic equation.
- Reject candidates outside the original logarithmic domain.
- Verify in the original equation when useful.
Common Errors
Equating Exponents with Different Bases
From
\[ 2^x=3^5, \]we cannot conclude
\[ x=5. \]The bases must match before the one-to-one exponential property can be used this way.
Forgetting to Isolate the Exponential Term
In
\[ 3\cdot2^x+1=25, \]first isolate \(2^x\).
Using False Logarithm Laws
In general,
\[ \log(a+b)\ne\log a+\log b. \]Ignoring Logarithmic Domains
Every real logarithm requires a positive argument.
Accepting Every Algebraic Candidate
A candidate produced after combining logarithms may violate the original domain.
Forcing a Closed Form When None Is Available
Some equations require numerical or graphical methods rather than elementary symbolic isolation.
Practice
1. Common Base
Solve:
\[ 2^x=32. \]Check Answer
\[ 32=2^5, \]so
\[ x=5. \]2. Rewrite Both Bases
Solve:
\[ 9^x=27. \]Check Answer
\[ 9=3^2, \qquad 27=3^3. \]So
\[ 3^{2x}=3^3, \]hence
\[ x=\frac32. \]3. Use Logarithms
Solve:
\[ 3^x=10. \]Check Answer
\[ x=\frac{\ln10}{\ln3}. \]4. Basic Logarithm
Solve:
\[ \log_4 x=3. \]Check Answer
\[ x=4^3=64. \]5. Shifted Logarithm
Solve:
\[ \ln(x-2)=0. \]Check Answer
Domain:
\[ x>2. \]Since
\[ e^0=1, \]we have
\[ x-2=1, \]so
\[ x=3. \]6. Combine Logarithms
Solve:
\[ \log x+\log(x-9)=1. \]Check Answer
Domain:
\[ x>9. \]Combine:
\[ \log[x(x-9)]=1. \]Therefore,
\[ x(x-9)=10. \]So
\[ x^2-9x-10=0. \]Factor:
\[ (x-10)(x+1)=0. \]The candidates are \(10\) and \(-1\), but only
\[ x=10 \]satisfies the domain.
7. Repeated Exponential Structure
Solve:
\[ 4^x-5\cdot2^x+4=0. \]Check Answer
Since
\[ 4^x=(2^x)^2, \]let
\[ u=2^x. \]Then
\[ u^2-5u+4=0, \] \[ (u-1)(u-4)=0. \]Thus,
\[ 2^x=1 \]or
\[ 2^x=4. \]Therefore,
\[ x=0 \]or
\[ x=2. \]Key Ideas
- Exponential equations place the variable in an exponent; logarithmic equations place it inside a logarithm.
- Exponential and logarithmic functions are inverse functions.
- When both sides can be rewritten with the same valid exponential base, equate the exponents.
- When a common base is inconvenient, logarithms can bring the exponent down as a coefficient.
- Real logarithmic arguments must always be positive.
- Equal logarithms with the same base have equal arguments when both arguments are in the domain.
- Logarithm laws apply to products, quotients, and powers—not to sums inside a logarithm.
- Substitution can simplify equations with repeated exponential structure.
- Algebraic candidates from logarithmic equations must be checked against the original domain.
- Exponential functions with positive bases have positive outputs, which can rule out some equations immediately.
- Growth, decay, doubling-time, and half-life equations naturally lead to logarithmic solving methods.
- Some equations mixing exponential and ordinary algebraic terms require numerical or graphical methods instead of elementary closed forms.
Where This Leads
Exponential and logarithmic equations extend inverse-operation reasoning into functions whose variables occur in fundamentally different positions.
They also prepare the way for exponential models, compound growth, logarithmic scales, inverse functions, numerical methods, and more advanced equations in which symbolic and graphical reasoning work together.