Mathematics 10 — Exercise 5.4 · Q1 (vi)

Full question

Solve the following equation by reducing it to a quadratic equation. Check the resulting values in the original equation.

Step-by-step solution

1. Use the exponent laws to write the equation as a quadratic in t, where t is the following positive power.

   $$t=3^x$$

2. After substitution the quadratic equation is:

   $$t^2-12\times t+27=0$$

3. Substitute the function expressions, keeping the brackets.

   $$t^2-12\times t+27-0$$

4. Remove the brackets. A preceding minus sign reverses every sign inside that bracket.

   $$t^2-12\times t+27$$

5. Write EVERY distributed product before collecting terms.

   $$27+t^2-12\times t-0$$

6. Calculate this individual product, including its sign and powers.

   $$27=27$$

7. Calculate this individual product, including its sign and powers.

   $$t^2=t^2$$

8. Calculate this individual product, including its sign and powers.

   $$-12\times t=-12\times t$$

9. Calculate this individual product, including its sign and powers.

   $$\left(-1\right)\times 0=0$$

10. These terms have the same variable part. Add ONLY their coefficients.

    $$27+0=27$$

11. These terms have the same variable part. Add ONLY their coefficients.

    $$1=1$$

12. These terms have the same variable part. Add ONLY their coefficients.

    $$-12=-12$$

13. Put the simplified groups together.

    $$t^2-12\times t+27$$

14. Collect terms with the same power and simplify their coefficients.

    $$t^2-12\times t+27$$

15. Expand the brackets and subtract the right-hand terms from both sides. Collect powers of the variable.

    $$t^2-12\times t+27=0$$

16. Split the middle coefficient into two numbers whose sum is b and whose product is a·c.

    $$(-3-9=-12,\left(-3\right)\times \left(-9\right)=27)$$

17. Rewrite the middle term using those two parts.

    $$t^2-3\times t-9\times t+27=0$$

18. Group the first two terms and the last two. Take out the common factor from each group.

    $$t\times \left(t-3\right)-9\times \left(t-3\right)=0$$

19. The common bracket can now be factored out.

    $$\left(t-9\right)\times \left(t-3\right)=0$$

20. A product is zero only if at least one factor is zero. Set each linear factor equal to zero.

    $$(t-9=0,t-3=0)$$

21. Take this factor equal to zero.

    $$t-9=0$$

22. Subtract the constant from both sides.

    $$t-9+9=0+9$$

23. Simplify both sides.

    $$t=9$$

24. Divide both sides by the coefficient of the variable.

    $$t=9\times \frac{1}{1}$$

25. Reduce the fraction to obtain this root.

    $$t=9$$

26. Take this factor equal to zero.

    $$t-3=0$$

27. Subtract the constant from both sides.

    $$t-3+3=0+3$$

28. Simplify both sides.

    $$t=3$$

29. Divide both sides by the coefficient of the variable.

    $$t=3\times \frac{1}{1}$$

30. Reduce the fraction to obtain this root.

    $$t=3$$

31. Substitute this root into the original equation before simplifying.

    $$0=0$$

32. Calculate both sides; their equal values confirm the root.

    $$0=0$$

33. Substitute this root into the original equation before simplifying.

    $$0=0$$

34. Calculate both sides; their equal values confirm the root.

    $$0=0$$

35. Set the original power equal to this positive t value.

    $$3^x=3$$

36. Write both sides with the same base if possible; otherwise use a logarithm.

    $$x=1$$

37. Divide the exponent equation by the multiplier of x.

    $$x=1$$

38. Set the original power equal to this positive t value.

    $$3^x=9$$

39. Write both sides with the same base if possible; otherwise use a logarithm.

    $$x=\frac{\log \left(9\right)}{\log \left(3\right)}$$

40. Divide the exponent equation by the multiplier of x.

    $$x=2$$

41. Collect every valid root from every substitution branch.

    $$\{1,2\}$$

Answer

The final answer is shown below.