No, it has a solution. The four rules of logarithms are: 1.2. 3. 4. log equation is equivalent to exponential equation 5. log2(x-3) + log2(x) = 2 + log2(x+2) Use rule 1 on the left side: log2[(x-3)x] = 2 + log2(x+2) Distribute log2(x²-3x) = 2 + log2(x+2) Get bothlogs on the left side of the equation: log2(x²-3x) - log2(x+2) = 2 Use rule 2 on the left log2 = 2 Use rule 4 = 22 Write 2² as 4 = 4 Multiply both sides by x+2 x² - 3x = 4(x + 2) x² - 3x = 4x + 8 x² - 7x - 8 = 0 (x - 8)(x + 1) = 0 x - 8 = 0; x + 1 = 0 x = 8; x = -1 The only solution is x = 8. Logarithms of negative numbers are not real numbers. So we discard the negative answer. Checking: log2(x-3) + log2(x) = 2 + log2(x+2) log2(8-3) + log2(8) = 2 + log2(8+2) log2(5) + log2(2³) = 2 + log2(10) Use rule 3 on 2nd term, Write 10 as 2·5 log2(5) + 3·log2(2) = 2 + log2(2·5) Use rule 5 to rewrite log2(2) as 1 Use rule 1 to rewrite last term on right log2(5) + 3·1 = 2 + log2(2) + log2(5) Use rule 5 to rewrite log2(2) as 1 log2(5) + 3 = 2 + 1 + log2(5) Add 2 + 1 log2(5) + 3 = 3 + log2(5) So it checks. Edwin