Move all the terms containing a logarithm to the left side of the equation.log(2x+1)−log(x−2)=1log(2x+1)-log(x-2)=1Use the quotient property of logarithms, logb(x)−logb(y)=logb(xy)logb(x)-logb(y)=logb(xy).log(2x+1x−2)=1log(2x+1x-2)=1Rewrite log(2x+1x−2)=1log(2x+1x-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)=ylogb(x)=y is equivalent to by=xby=x.101=2x+1x−2101=2x+1x-2Solve for xxTap for more steps...x=218x=218Verify 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=218x=218The result can be shown in both exact and decimal forms.Exact Form:x=218x=218Decimal Form:x=2.625