How do you solve log(x) = k?

Convert to exponential form using the definition: log_a(x) = k means x = a^k. For example, log₂(x) = 5 gives x = 2⁵ = 32. For ln(x) = 4, x = e⁴ ≈ 54.6.

How do you solve equations with log on both sides?

Use the one-to-one property: if log_a(M) = log_a(N), then M = N. For log₄(3x+2) = log₄(x+10), set 3x+2 = x+10 and solve to get x = 4.

How do you combine logarithms to solve an equation?

Use logarithm rules to condense multiple logs into one. For log₂(x) + log₂(x-2) = 3, combine to log₂(x(x-2)) = 3, then convert: x² - 2x = 8.

What are domain restrictions in logarithmic equations?

Every argument of every logarithm must be positive. For log(x-5) + log(x+1), you need x-5 > 0 AND x+1 > 0, so x > 5. Check all solutions against these restrictions.

What are extraneous solutions in log equations?

Extraneous solutions are algebraic solutions that violate domain restrictions. For log(x) + log(x-3) = 1, solving gives x = 5 or x = -2, but x = -2 makes log(-2) undefined, so reject it.

How do you solve (log x)² - 5log(x) + 6 = 0?

Use substitution: let u = log(x). Solve u² - 5u + 6 = 0 to get u = 2 or u = 3. Back-substitute: log(x) = 2 gives x = 100, log(x) = 3 gives x = 1000.

How do you solve exponential equations like 3^x = 7?

Take logarithms of both sides: log(3^x) = log(7), so x·log(3) = log(7). Divide: x = log(7)/log(3) ≈ 1.771. Either log or ln gives the same answer.

How do you solve 2^(x+3) = 5^(x-1)?

Take ln of both sides: (x+3)ln(2) = (x-1)ln(5). Expand, collect x terms on one side, and solve. This gives x = (ln5 + 3ln2)/(ln5 - ln2) ≈ 4.03.

Does it matter if I use log or ln to solve equations?

No. For 3^x = 7, both log(7)/log(3) and ln(7)/ln(3) give the same decimal answer. Use ln when the base is e, log when base is 10, otherwise either works.

Why must you always check solutions in log equations?

Algebraic manipulation can produce values that make original logarithms undefined. Substituting back into the original equation catches these extraneous solutions before reporting the final answer.

Logarithmic Equations

Basic Form: Logarithm Equals a Constant

Logarithms on Both Sides

Combining Logarithms

Domain Restrictions

Extraneous Solutions

Equations Requiring Substitution

Solving Exponential Equations via Logarithms

Exponentials on Both Sides

Choosing Common or Natural Logarithm

Isolating the Variable Inside the Logarithm

Equations containing logarithms require techniques that exploit the definition and properties of logarithmic functions. Converting to exponential form, applying the one-to-one property, and using logarithm rules to combine expressions — these methods reduce logarithmic equations to algebraic ones. Every solution must be verified against domain restrictions, as the solving process can introduce extraneous values.

Basic Form: Logarithm Equals a Constant

The simplest logarithmic equation has the form

\log_a(x) = k

, where

k

is a constant.

Convert to exponential form using the definition:

\log_a(x) = k

means

a^k = x

.

Example: Solve

\log_2(x) = 5

x = 2^5 = 32

Example: Solve

\log_3(x) = -2

x = 3^{-2} = \frac{1}{9}

Example: Solve

\ln(x) = 4

x = e^4 \approx 54.598

When the argument is an expression, solve the resulting equation:

Example: Solve

\log_5(2x - 1) = 2

2x - 1 = 5^2 = 25

2x = 26

x = 13

Logarithms on Both Sides

When the same logarithm base appears on both sides, apply the one-to-one property: if

\log_a(M) = \log_a(N)

, then

M = N

.

Example: Solve

\log_4(3x + 2) = \log_4(x + 10)

3x + 2 = x + 10

2x = 8

x = 4

Verification:

\log_4(3(4) + 2) = \log_4(14)

and

\log_4(4 + 10) = \log_4(14)

. Both arguments are positive, and the equation holds.

Example: Solve

\ln(x^2) = \ln(9)

x^2 = 9

x = \pm 3

Check both:

\ln((-3)^2) = \ln(9)

works, and

\ln((3)^2) = \ln(9)

works. Both solutions are valid.

Combining Logarithms

When multiple logarithms appear, use the rules to condense them into a single logarithm, then apply previous techniques.

Example: Solve

\log_2(x) + \log_2(x - 2) = 3

.

Condense using the product rule:

\log_2(x(x - 2)) = 3

\log_2(x^2 - 2x) = 3

Convert to exponential form:

x^2 - 2x = 2^3 = 8

x^2 - 2x - 8 = 0

(x - 4)(x + 2) = 0

x = 4 \text{ or } x = -2

Check domain:

x = 4

gives

\log_2(4) + \log_2(2)

, both defined.

x = -2

gives

\log_2(-2)

, undefined. Reject

x = -2

.

Solution:

x = 4

Domain Restrictions

Every argument of every logarithm in the original equation must be positive. Identify these restrictions before solving.

For

\log_a(f(x))

to be defined,

f(x) > 0

.

Example: In

\log_3(x - 5) + \log_3(x + 1) = 2

, the restrictions are:

x - 5 > 0 \implies x > 5

x + 1 > 0 \implies x > -1

The intersection is

x > 5

. Any solution must satisfy

x > 5

.

Solving:

\log_3((x-5)(x+1)) = 2

(x-5)(x+1) = 9

x^2 - 4x - 5 = 9

x^2 - 4x - 14 = 0

x = \frac{4 \pm \sqrt{16 + 56}}{2} = \frac{4 \pm \sqrt{72}}{2} = 2 \pm 3\sqrt{2}

Since

3\sqrt{2} \approx 4.24

, we have

x \approx 6.24

x \approx -2.24

. Only

x = 2 + 3\sqrt{2}

satisfies

x > 5

Extraneous Solutions

Extraneous solutions arise when algebraic manipulation produces values that violate the original equation's domain. Always substitute back into the original equation.

Example: Solve

\log(x) + \log(x - 3) = 1

\log(x(x-3)) = 1

x(x-3) = 10

x^2 - 3x - 10 = 0

(x-5)(x+2) = 0

x = 5 \text{ or } x = -2

Check

x = 5

\log(5) + \log(2)

— both defined, sum equals

\log(10) = 1

. Valid.

Check

x = -2

\log(-2)

— undefined. Extraneous.

The squaring, factoring, or combining steps do not "create" these solutions — they were always algebraic solutions to the transformed equation. The domain restrictions eliminate them.

Equations Requiring Substitution

Some equations have a quadratic structure in the logarithm. Substitution simplifies the algebra.

Example: Solve

(\log_2(x))^2 - 5\log_2(x) + 6 = 0

.

Let

u = \log_2(x)

u^2 - 5u + 6 = 0

(u - 2)(u - 3) = 0

u = 2 \text{ or } u = 3

Back-substitute:

\log_2(x) = 2 \implies x = 4

\log_2(x) = 3 \implies x = 8

Both satisfy

x > 0

. Solutions:

x = 4

and

x = 8

.

Example: Solve

\ln(x) + \frac{6}{\ln(x)} = 5

.

Let

u = \ln(x)

u + \frac{6}{u} = 5

u^2 + 6 = 5u

u^2 - 5u + 6 = 0

(u-2)(u-3) = 0

Back-substitute:

x = e^2

x = e^3

Solving Exponential Equations via Logarithms

When an exponential equation cannot be solved by matching bases, take the logarithm of both sides.

Example: Solve

3^x = 7

.

No integer power of

3

equals

7

. Take logarithms:

\log(3^x) = \log(7)

x \cdot \log(3) = \log(7)

x = \frac{\log(7)}{\log(3)} \approx \frac{0.845}{0.477} \approx 1.771

Either

\log

\ln

works — the ratio is identical.

Example: Solve

5^{2x+1} = 12

\ln(5^{2x+1}) = \ln(12)

(2x+1)\ln(5) = \ln(12)

2x + 1 = \frac{\ln(12)}{\ln(5)}

2x = \frac{\ln(12)}{\ln(5)} - 1

x = \frac{1}{2}\left(\frac{\ln(12)}{\ln(5)} - 1\right) \approx 0.271

Exponentials on Both Sides

When exponential expressions with different bases appear on both sides, logarithms bring down both exponents.

Example: Solve

2^{x+3} = 5^{x-1}

.

Take logarithms:

\ln(2^{x+3}) = \ln(5^{x-1})

(x+3)\ln(2) = (x-1)\ln(5)

Expand:

x\ln(2) + 3\ln(2) = x\ln(5) - \ln(5)

Collect

x

terms:

x\ln(2) - x\ln(5) = -\ln(5) - 3\ln(2)

x(\ln(2) - \ln(5)) = -\ln(5) - 3\ln(2)

x = \frac{-\ln(5) - 3\ln(2)}{\ln(2) - \ln(5)} = \frac{\ln(5) + 3\ln(2)}{\ln(5) - \ln(2)}

Numerically:

x = \frac{1.609 + 2.079}{1.609 - 0.693} = \frac{3.688}{0.916} \approx 4.026

Choosing Common or Natural Logarithm

When taking logarithms to solve exponential equations, either base produces the same answer.

For

3^x = 7

x = \frac{\log(7)}{\log(3)} = \frac{\ln(7)}{\ln(3)} = \log_3(7)

All three expressions are equal by the change of base formula.

Practical considerations:

$e$ , use $\ln$ : solving $e^{2x} = 5$ gives $2x = \ln(5)$ , no division needed.

$10$ , use $\log$ : solving $10^{x-1} = 50$ gives $x - 1 = \log(50)$ .

The expressions

\log(7)/\log(3)

and

\ln(7)/\ln(3)

compute to identical decimal values.