Chalk−1

Math · College algebra · Worked example

Solve a logarithmic equation and check it

Solve log₂(x + 1) + log₂(x − 1) = 3.

log2(x+1)+log2(x−1)=3

Combine the logarithms

The product rule turns the sum of logs into the log of a product: (x + 1)(x − 1) = x² − 1.

log2(x2−1)=3

Rewrite in exponential form

log₂(x² − 1) = 3 means x² − 1 = 2³ = 8.

x2−1=8

Solve the quadratic

x² = 9, so the candidates are x = 3 and x = −3.

x=3 or x=−3

Check each candidate

x = 3 gives log₂ 4 + log₂ 2 = 2 + 1 = 3, which works. x = −3 gives log₂(−2), which is undefined, so it is rejected. Combining the logs widened the domain, which is how the extra candidate crept in.

Result

x = 3. The candidate x = −3 is extraneous.

x=3

Your turn

Solve ln x + ln(x − 3) = ln 10.

Show the answer and explanation

x = 5.

Combine: ln(x² − 3x) = ln 10, so x² − 3x − 10 = 0 and (x − 5)(x + 2) = 0. x = −2 makes ln x undefined, so only x = 5 works: ln 5 + ln 2 = ln 10.

Keep exploring

Open the graph: the curve y = log₂(x + 1) + log₂(x − 1) exists only for x > 1, and it meets y = 3 once, at x = 3.

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