Exponential and logarithmic equations

An exponential equation has the unknown in an exponent, such as (3^{2x-1}=17). A logarithmic equation contains a logarithm of an expression involving the unknown, such as (log_2(x-1)=3).

The main idea is to use inverse operations:

[ a^x=y \quad\Longleftrightarrow\quad x=\log_a y, ]

where (a>0), (a\ne1) and (y>0). Logarithms undo exponentials, just as square roots undo squaring.

Prerequisites

You should be able to:

• use the laws of indices
• use logarithms and their laws
• solve quadratic equations
• use composite and inverse functions

Choose a method before calculating

For an exponential equation, try these methods in order:

1. Rewrite both sides with the same base.
2. Factor out a common exponential expression.
3. Substitute (u=a^x) if powers such as (a^{2x}) and (a^x) occur together.
4. If none of these works, take logarithms.

For a logarithmic equation:

1. State the domain restrictions.
2. Combine logarithms where useful.
3. Convert between logarithmic and exponential form.
4. Solve the resulting algebraic equation.
5. Check every answer in the original equation.

Exponential equations with a common base

If (a>0) and (a\ne1), the function (a^x) is one-to-one. Therefore

[ a^p=a^q \quad\Longrightarrow\quad p=q. ]

Example 1: rewrite both sides

Solve

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

Write both sides as powers of (2):

[ (2^3)^{x-1}=(2^2)^{2x+1}. ]

Apply ((a^m)^n=a^{mn}):

[ 2^{3x-3}=2^{4x+2}. ]

Equate the exponents:

[ 3x-3=4x+2, ]

so

[ \boxed{x=-5}. ]

Example 2: recognise a reciprocal base

Solve

[ 9^{x+1}=\left(\frac13\right)^{2x-4}. ]

Since (9=3^2) and (1/3=3^{-1}),

[ 3^{2x+2}=3^{-2x+4}. ]

Thus

[ 2x+2=-2x+4, ]

and

[ \boxed{x=\frac12}. ]

Do not equate exponents too early

From (2^p=3^q), it does not follow that (p=q). Exponents may be equated only when the positive bases are the same.

Taking logarithms

When the two sides cannot sensibly be written with one base, take (ln) or (log) of both sides. Either logarithm gives the same result.

Example 3: an exact answer and a decimal answer

Solve

[ 5^{2x-1}=19. ]

Take natural logarithms:

[ \ln\left(5^{2x-1}\right)=\ln 19. ]

Using (ln(a^k)=k\ln a),

[ (2x-1)\ln5=\ln19. ]

Therefore

[ x=\frac12\left(1+\frac{\ln19}{\ln5}\right) \approx1.415. ]

Keep the logarithmic form during the working. Round only the final answer.

Example 4: exponentials on both sides

Solve

[ 2^{x+1}=7^{3-x}. ]

Take logarithms:

[ (x+1)\ln2=(3-x)\ln7. ]

Collect the (x)-terms:

[ x\ln2+x\ln7=3\ln7-\ln2. ]

Hence

[ \boxed{x=\frac{3\ln7-\ln2}{\ln2+\ln7}} \approx1.949. ]

Why taking logarithms works

Both sides of an exponential equation are positive, so their logarithms exist. Since (ln x) is one-to-one on (x>0), applying (ln) to both sides preserves equivalence.

Do not confuse these statements:

[ \ln(5^x)=x\ln5 ]

is correct, but

[ \ln(5^x)\ne(\ln5)^x. ]

Factorising exponential expressions

Index laws can reveal a common factor.

Example 5: factor out the smaller power

Solve

[ 3^{x+1}-3^x=54. ]

Since (3^{x+1}=3\cdot3^x),

[ 3\cdot3^x-3^x=54. ]

Factorise:

[ 3^x(3-1)=54. ]

Thus

[ 3^x=27=3^3, ]

so

[ \boxed{x=3}. ]

Example 6: different shifts in the exponent

Solve

[ 2^{x+2}+2^{x-1}=36. ]

Factor out (2^{x-1}). Since (2^{x+2}=2^{x-1}2^3),

[ 2^{x-1}(8+1)=36. ]

Therefore

[ 2^{x-1}=4=2^2, ]

giving

[ \boxed{x=3}. ]

Equations that are quadratic in an exponential

Because (a^{2x}=(a^x)^2), an equation involving (a^{2x}) and (a^x) may be a disguised quadratic.

Example 7: substitution and rejection

Solve

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

Let

[ u=2^x. ]

Then (4^x=(2^x)^2=u^2), so

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

Factorise:

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

Hence (u=1) or (u=4). Substituting back,

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

or

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

Therefore

[ \boxed{x=0\text{ or }x=2}. ]

Example 8: the substitution has a restricted range

Solve

[ e^{2x}+2e^x-8=0. ]

Let (u=e^x). Crucially, (u>0). Then

[ u^2+2u-8=0, ]

so

[ (u+4)(u-2)=0. ]

The algebra gives (u=-4) or (u=2). Reject (u=-4) because (e^x) is always positive. Thus

[ e^x=2 ]

and

[ \boxed{x=\ln2}. ]

Logarithmic equations and their domains

For real logarithms, every logarithm argument must be strictly positive:

[ \log_a(f(x))\text{ is defined only when }f(x)>0. ]

This is why logarithmic equations can produce algebraic candidates that must be rejected.

Example 9: convert to exponential form

Solve

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

The domain condition is

[ 2x-1>0 \Rightarrow x>\frac12. ]

Convert to exponential form:

[ 2x-1=3^2=9. ]

Therefore

[ \boxed{x=5}, ]

which satisfies the domain condition.

Example 10: combine logarithms first

Solve

[ \ln x+\ln(x-2)=\ln8. ]

The arguments require

[ x>0\quad\text{and}\quad x-2>0, ]

so the combined domain is (x>2).

Use (ln p+\ln q=\ln(pq)):

[ \ln\bigl(x(x-2)\bigr)=\ln8. ]

Because (ln) is one-to-one,

[ x(x-2)=8. ]

Thus

[ x^2-2x-8=0, ]

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

The candidates are (x=4) and (x=-2). Only (x=4) lies in (x>2), so

[ \boxed{x=4}. ]

Example 11: subtraction of logarithms

Solve

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

The domain is (x>1). Apply the quotient law:

[ \log_2\left(\frac{x+3}{x-1}\right)=2. ]

Convert to exponential form:

[ \frac{x+3}{x-1}=2^2=4. ]

Hence

[ x+3=4x-4, ]

so

[ \boxed{x=\frac73}. ]

Since (7/3>1), this is valid.

Example 12: different logarithm bases

Solve

[ \log_2 x+\log_4 x=6. ]

Use the change of base formula:

[ \log_4x=\frac{\log_2x}{\log_24}=\frac12\log_2x. ]

Therefore

[ \frac32\log_2x=6, ]

so

[ \log_2x=4 ]

and

[ \boxed{x=16}. ]

Equations without an elementary exact solution

Some equations mix (x) and an exponential in a way that cannot be rearranged using standard A-level functions. For example,

[ e^x=3x. ]

Taking logarithms gives (x=\ln(3x)), but this still contains (x) on both sides. Such an equation should be solved graphically or with a numerical method. It actually has two positive roots, approximately

[ x\approx0.619\quad\text{and}\quad x\approx1.512. ]

A calculator graph or a sign-change search helps locate all roots before iteration. Continue with change of sign, fixed-point iteration and Newton-Raphson.

Common misconceptions

• Splitting a logarithm of a sum: (ln(a+b)\ne\ln a+\ln b). The logarithm laws apply to products, quotients and powers.
• Forgetting positivity: (a^x>0), so an equation such as (2^x=-3) has no real solution.
• Ignoring the logarithm domain: (ln(x-4)) requires (x>4), not (x\ge4).
• Taking logs term by term: (ln(2^x+1)) cannot be simplified to (x\ln2+\ln1).
• Rounding too soon: retain exact logarithms until the final line.
• Finding only one numerical root: inspect the graph or intervals before concluding that all roots have been found.

Self-check

Try these without looking back at the examples.

1. Solve (27^{x-1}=9^{2x+1}).
2. Solve (4^x=11), giving an exact answer and a decimal answer to three decimal places.
3. Solve (5^{x+1}-5^x=100).
4. Solve (9^x-10(3^x)+9=0).
5. Solve (ln(2x+1)=3).
6. Solve (log_3x+\log_3(x-2)=1), checking the domain.

Answers

1. (3^{3x-3}=3^{4x+2}), so (x=-5).
2. (x=\dfrac{\ln11}{\ln4}\approx1.730).
3. (5^x(5-1)=100), so (5^x=25) and (x=2).
4. Let (u=3^x>0). Then ((u-1)(u-9)=0), giving (x=0) or (x=2).
5. (2x+1=e^3), so (x=\dfrac{e^3-1}{2}).
6. The domain is (x>2). Then (x(x-2)=3), so ((x-3)(x+1)=0). Reject (-1), giving (x=3).

Summary

The equation’s structure determines the method:

[ \begin{array}{c|c} \text{Structure} & \text{Useful method}\ \hline a^{f(x)}=a^{g(x)} & f(x)=g(x)\ a^{f(x)}=k & f(x)=\dfrac{\ln k}{\ln a}\ a^{2x}+pa^x+q=0 & u=a^x,\ u>0\ \log_a(f(x))=k & f(x)=a^k\ \log_a p+\log_a q & \log_a(pq)\ \log_a p-\log_a q & \log_a(p/q) \end{array} ]

For every logarithmic equation, impose positive arguments and check solutions in the original equation. For exponential substitutions, remember that the substituted quantity is positive. These checks are part of the solution, not optional finishing touches.

Next, apply these techniques to exponential growth and decay and to linearising exponential data.