Combine logarithms only after stating their common domain.
Notes and key methods
In a logarithmic equation every original argument must be positive. Combine logarithms using a valid law, exponentiate in the given base, and solve the resulting algebraic equation.
The algebraic equation may also have roots outside the original domain. Reject those explicitly and substitute each retained root into the original arguments.
Check the conditions
Every original logarithm argument is strictly positive.
The base is positive and is not one.
Three worked examples
Example 1 · Logarithmic equations and domains
Solve \(\log_{4}(x - 3)+\log_{4}(x)=\log_{4}(28)\). Here \(\displaystyle h=3\).
State the domain and solve the equation completely.
Find the rejected algebraic root.
Find x−h at the valid solution.
Find the product of the two logarithm arguments at the valid solution.
Show the reasoning
(a)
\(x \left(x - 3\right) = 28\) gives \(\displaystyle x=7\) or -4. The original arguments require x>3, so only 7 is valid.
Answer: \(7\) with \(x > 3\)
(b)
Check the original arguments.
\(-4\) makes x−h negative.
Answer: \(-4\).
(c)
Substitute the valid x-value.
\(4\)
Answer: \(4\).
(d)
Verify the combined logarithm equation.
\(28\)
Answer: \(28\).
Example 2 · Logarithmic equations and domains
Solve \(\log_{2}(x - 4)+\log_{2}(x - 3)=\log_{2}(30)\). Here \(\displaystyle h=4\).
State the domain and solve the equation completely.
Find the rejected algebraic root.
Find x−h at the valid solution.
Find the product of the two logarithm arguments at the valid solution.
Show the reasoning
(a)
\(\left(x - 4\right) \left(x - 3\right) = 30\) gives \(\displaystyle x=9\) or -2. The original arguments require x>4, so only 9 is valid.
Answer: \(9\) with \(x > 4\)
(b)
Check the original arguments.
\(-2\) makes x−h negative.
Answer: \(-2\).
(c)
Substitute the valid x-value.
\(5\)
Answer: \(5\).
(d)
Verify the combined logarithm equation.
\(30\)
Answer: \(30\).
Example 3 · Logarithmic equations and domains
Solve \(\log_{3}(x - 5)+\log_{3}(x - 3)=\log_{3}(8)\). Here \(\displaystyle h=5\).
State the domain and solve the equation completely.
Find the rejected algebraic root.
Find x−h at the valid solution.
Find the product of the two logarithm arguments at the valid solution.
Show the reasoning
(a)
\(\left(x - 5\right) \left(x - 3\right) = 8\) gives \(\displaystyle x=7\) or 1. The original arguments require x>5, so only 7 is valid.
Answer: \(7\) with \(x > 5\)
(b)
Check the original arguments.
\(1\) makes x−h negative.
Answer: \(1\).
(c)
Substitute the valid x-value.
\(2\)
Answer: \(2\).
(d)
Verify the combined logarithm equation.
\(8\)
Answer: \(8\).
Common mistakes
Exponentiating does not make a negative logarithm argument valid.
Check the original domain, not just the combined argument.