Check→43/2 = 8, which confirms that the value is 3/2.
Example 5: an equation with two logarithms of the same base
- Solve log2 x + log2 (x − 2) = 3.
- Solution By the product law, log2 (x(x − 2)) = 3, so x(x − 2) = 23 = 8. This gives x2 − 2x − 8 = 0, and factorising (x − 4)(x + 2) = 0 gives x = 4 or x = −2.
- The domain requires x > 0 and x − 2 > 0, that is x > 2. So we reject −2 and keep x = 4.
- Check log2 4 + log2 2 = 2 + 1 = 3. The solution is correct.
Example 6: an equation with a difference of two logarithms
- Solve log3 (x + 1) − log3 (x − 1) = 1.
- Solution By the quotient law, log3 ((x + 1) / (x − 1)) = 1, so (x + 1) / (x − 1) = 3. Multiplying by (x − 1) gives x + 1 = 3x − 3, so 2x = 4 and x = 2.