Question

Solve the equation algebraically for all values of X

Answer 1

This is a problem you need to solve using logs. When you use logs you can "pull" the exponents down in front of the log to get a new equation that looks like this: 2x^3 + x^2 log 81 = 6x - 3 log 27. Now divide both sides by log 81 and 6x - 3 simultaneously to get (2x^3 + x^2)/(6x - 3) = (log 27)/(log 81). If you do the log math on the right side you get .75. Now multiply both sides by 6x-3 to get 2x^3+x^2 = .75(6x-3). If you distribute that out on the left side you'll get 2x^3+x^2=4.5x-2.25. Now move everything over to the left side and set the whole thing equal to 0: 2x^3+x^2-4.5x+2.25=0. When you solve for x, you are in essence factoring, so do this by grouping: x^2(2x+1)-2.25(2x+1). Now finally factor out the 2x+1 to get (2x+1)(x^2-2.25). You're not done yet though cuz you need to solve each of those for x: 2x+1=0, and x= -1/2; x^2=2.25, and x=+/- 1.5. So all the values for x here are -1/2, 1.5, and -1.5