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About this lesson Algebra II › Logarithms
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Solve logarithmic equations and check the domain Logarithms. log rules, change of base, logarithmic equations, half-life
Warm-up Three quick questions from earlier lessons.
Discover Solve log 2 x = 3 \log_2 x = 3 log 2 x = 3 by rewriting it as a power.
Then think: can ln x \ln x ln x ever be defined for x = − 2 x = -2 x = − 2 ? What does that mean for solutions you find?
Key idea To solve a logarithmic equation: combine into one logarithm, rewrite as a power (or compare arguments), solve, then check the domain .
log b ( expression ) = n ⟺ expression = b n \log_b(\text{expression}) = n \iff \text{expression} = b^{n} log b ( expression ) = n ⟺ expression = b n , and every argument of a logarithm must be positive
Worked example Solve ln x + ln ( x − 2 ) = ln 15 \ln x + \ln(x - 2) = \ln 15 ln x + ln ( x − 2 ) = ln 15 .
ln ( x ( x − 2 ) ) = ln 15 \ln\big(x(x - 2)\big) = \ln 15 ln ( x ( x − 2 ) ) = ln 15 Product rule.
x 2 − 2 x − 15 = 0 x^2 - 2x - 15 = 0 x 2 − 2 x − 15 = 0 Equal logarithms have equal arguments.
( x − 5 ) ( x + 3 ) = 0 (x - 5)(x + 3) = 0 ( x − 5 ) ( x + 3 ) = 0 Factor.
x = −3 makes ln x undefined: reject it.
Your turn, guided Your turn: solve log 2 ( x − 3 ) = 4 \log_2(x - 3) = 4 log 2 ( x − 3 ) = 4 . Fill in the missing line.
x − 3 = 2 4 = 16 x - 3 = 2^4 = 16 x − 3 = 2 4 = 16 Rewrite as a power.
Add 3. Check: 19 − 3 = 16 > 0.
Practice ★ Put the steps for solving ln x + ln ( x − 5 ) = ln 6 \ln x + \ln(x - 5) = \ln 6 ln x + ln ( x − 5 ) = ln 6 in order. Combine with the product rule: ln ( x ( x − 5 ) ) = ln 6 \ln\big(x(x - 5)\big) = \ln 6 ln ( x ( x − 5 ) ) = ln 6 Equal logarithms have equal arguments: x ( x − 5 ) = 6 x(x - 5) = 6 x ( x − 5 ) = 6 Solve the quadratic equation Keep only the values where x > 5 x > 5 x > 5 Put the steps for solving ln x + ln ( x − 4 ) = ln 21 \ln x + \ln(x - 4) = \ln 21 ln x + ln ( x − 4 ) = ln 21 in order. Combine with the product rule: ln ( x ( x − 4 ) ) = ln 21 \ln\big(x(x - 4)\big) = \ln 21 ln ( x ( x − 4 ) ) = ln 21 Equal logarithms have equal arguments: x ( x − 4 ) = 21 x(x - 4) = 21 x ( x − 4 ) = 21 Solve the quadratic equation Keep only the values where x > 4 x > 4 x > 4
Solve log 5 ( x − 8 ) = 4 \log_{ 5 }(x - 8) = 4 log 5 ( x − 8 ) = 4 . Solve log 5 ( x − 9 ) = 3 \log_{ 5 }(x - 9) = 3 log 5 ( x − 9 ) = 3 . Solve log 3 ( x − 8 ) = 4 \log_{ 3 }(x - 8) = 4 log 3 ( x − 8 ) = 4 . Solve log 2 ( x − 2 ) = 4 \log_{ 2 }(x - 2) = 4 log 2 ( x − 2 ) = 4 . Solve log 2 ( x − 6 ) = 4 \log_{ 2 }(x - 6) = 4 log 2 ( x − 6 ) = 4 .
Watch out! Lily solved ln x + ln ( x − 1 ) = ln 6 \ln x + \ln(x - 1) = \ln 6 ln x + ln ( x − 1 ) = ln 6 . Tap the line with the mistake.
x 2 − x − 6 = 0 x^2 - x - 6 = 0 x 2 − x − 6 = 0 Correct: product rule, then compare arguments.
x = 3 x = 3 x = 3 or
x = − 2 x = -2 x = − 2 Wrong: x = −2 makes ln x undefined, so it must be rejected. The only solution is x = 3.
Two learners solved ln x + ln ( x − 3 ) = ln 4 \ln x + \ln(x - 3) = \ln 4 ln x + ln ( x − 3 ) = ln 4 . Compare their work.
Solution A
x 2 − 3 x − 4 = 0 x^2 - 3x - 4 = 0 x 2 − 3 x − 4 = 0 Product rule, then compare the arguments.
( x − 4 ) ( x + 1 ) = 0 (x - 4)(x + 1) = 0 ( x − 4 ) ( x + 1 ) = 0 Factor.
x = 4 x = 4 x = 4 or
x = − 1 x = -1 x = − 1 Both values solve the quadratic.
Solution B
x 2 − 3 x − 4 = 0 x^2 - 3x - 4 = 0 x 2 − 3 x − 4 = 0 Product rule, then compare the arguments.
( x − 4 ) ( x + 1 ) = 0 (x - 4)(x + 1) = 0 ( x − 4 ) ( x + 1 ) = 0 Factor.
ln(−1) is not defined, so x = −1 is rejected.
Solution A is correct Solution B is correct Both are correct
Practice ★★ Solve ln x + ln ( x − 1 ) = ln 42 \ln x + \ln(x - 1) = \ln 42 ln x + ln ( x − 1 ) = ln 42 . Give every solution. Solve ln x + ln ( x − 2 ) = ln 24 \ln x + \ln(x - 2) = \ln 24 ln x + ln ( x − 2 ) = ln 24 . Give every solution. Solve ln x + ln ( x − 3 ) = ln 18 \ln x + \ln(x - 3) = \ln 18 ln x + ln ( x − 3 ) = ln 18 . Give every solution. Solve ln x + ln ( x − 4 ) = ln 45 \ln x + \ln(x - 4) = \ln 45 ln x + ln ( x − 4 ) = ln 45 . Give every solution. Solve ln x + ln ( x − 1 ) = ln 72 \ln x + \ln(x - 1) = \ln 72 ln x + ln ( x − 1 ) = ln 72 . Give every solution.
Exit ticket Three questions to check you have got it.
Solve ln x + ln ( x − 2 ) = ln 3 \ln x + \ln(x - 2) = \ln 3 ln x + ln ( x − 2 ) = ln 3 . Solve ln x + ln ( x − 2 ) = ln 48 \ln x + \ln(x - 2) = \ln 48 ln x + ln ( x − 2 ) = ln 48 . Solve ln x + ln ( x − 1 ) = ln 56 \ln x + \ln(x - 1) = \ln 56 ln x + ln ( x − 1 ) = ln 56 .
Summary card Solving logarithmic equations
Combine into one logarithm (product, quotient, power rules). Rewrite as a power, or compare arguments. Solve, then check that every argument is positive.
Answers: Solve logarithmic equations and check the domain A
1. Combine with the product rule: ln ( x ( x − 5 ) ) = ln 6 \ln\big(x(x - 5)\big) = \ln 6 ln ( x ( x − 5 ) ) = ln 6 2. Equal logarithms have equal arguments: x ( x − 5 ) = 6 x(x - 5) = 6 x ( x − 5 ) = 6 3. Solve the quadratic equation 4. Keep only the values where x > 5 x > 5 x > 5 1. Combine with the product rule: ln ( x ( x − 4 ) ) = ln 21 \ln\big(x(x - 4)\big) = \ln 21 ln ( x ( x − 4 ) ) = ln 21 2. Equal logarithms have equal arguments: x ( x − 4 ) = 21 x(x - 4) = 21 x ( x − 4 ) = 21 3. Solve the quadratic equation 4. Keep only the values where x > 4 x > 4 x > 4 B
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