Log (2x+1) = 1+log (x-2)

Move all the terms containing a logarithm to the left side of the equation. log (2x+1) - log (x-2) = 1 log (2x+1) - log (x-2) = 1 Use the quotient property of logarithms, logb (x) - logb (y) = logb (xy) logb (x) - logb (y) = logb (xy). log (2x+1 x-2) = 1 log (2x+1 x-2) = 1 Rewrite log (2x+1 x-2) = 1 log (2x+1 x-2) = 1 in exponential form using the definition of a logarithm. If xx and bb are positive real numbers and bb ≠≠ 11, then logb (x) = y logb (x) = y is equivalent to by = x by = x. 101 = 2x+1 x-2 101 = 2x+1 x-2 Solve for xx Tap for more steps ... x = 218 x = 218 Verify each of the solutions by substituting them back into the original equation log (2x+1) = 1+log (x-2) log (2x+1) = 1+log (x-2) and solving. In this case, all solutions were found to be valid. x = 218 x = 218 The result can be shown in both exact and decimal forms. Exact Form: x = 218 x = 218 Decimal Form: x = 2.625