Logarithmic equations questions with answers

Questions

1. Question 1One log

   Solve $\log_{2} (x + 2) = 2$.

   Show the worked solution

   1. Step 1: Write it in index form: logₐ N = k means N = aᵏ$x + 2 = 2^{2}$
   2. Step 2: Work out the power, then solve$x + 2 = 4$$x = 2$
   3. Step 3: Check: the log of a positive number only$(2) + 2 = 4 > 0$ ✓
2. Question 2Two logs added

   Solve $\lg x + \lg (x - 15) = 3$.

   Show the worked solution

   1. Step 1: Which law of logarithms combines the two logs into one?Product law: $\log_{a} m + \log_{a} n = \log_{a} mn$.
   2. Step 2: Use it to write a single log$\lg [x(x - 15)] = 3$
   3. Step 3: Index form: logₐ N = k means N = aᵏ$x(x - 15) = 10^{3}$$x^2 - 15x = 1000$$x^2 - 15x - 1000 = 0$
   4. Step 4: Factorise and solve the quadratic$(x - 40)(x + 25) = 0$$x = -25$ or $x = 40$
   5. Step 5: You can only take the log of a positive number: check each$x = -25$ is rejected ($\lg x$ has no value for $x \leq 0$).$x = 40$
3. Question 3Two logs subtracted

   Solve $\log_{3} (x + 7) - \log_{3} (x - 1) = 2$.

   Show the worked solution

   1. Step 1: Which law of logarithms combines the two logs into one?Quotient law: $\log_{a} m - \log_{a} n = \log_{a} \dfrac{m}{n}$.
   2. Step 2: Use it to write a single log$\log_{3} \dfrac{x + 7}{x - 1} = 2$
   3. Step 3: Index form, then multiply both sides by the bottom$\dfrac{x + 7}{x - 1} = 3^{2}$$x + 7$ $=$ $9(x - 1)$
   4. Step 4: Expand and solve$x + 7$ $=$ $9x - 9$$16 = 8x$$x = 2$
   5. Step 5: Check both logs are defined (positive inside)$(2) + 7 > 0$ and $(2) - 1 > 0$ ✓
4. Question 4Different bases: change of base

   Solve the equation $2\log_{4} x + \log_{2} x = 12$.

   Show the worked solution

   1. Step 1: Change of base: write log₄ x with base 2$\log_{4} x = \dfrac{\log_{2} x}{\log_{2} 4}$$= \dfrac{\log_{2} x}{2}$
   2. Step 2: Collect the terms in log₂ x$\log_{2} x + \dfrac{2\log_{2} x}{2} = 12$$\log_{2} x = 6$
   3. Step 3: Write the log as a power$x = 64$
5. Question 5A quadratic in logₐ x

   Solve the equation $\lg x - 2\log_{x} 10 = 1$.

   Show the worked solution

   1. Step 1: Change of base: logₓ 10 = 1 ÷ lg x. Let u = lg x$\log_{x} 10 = \dfrac{1}{\lg x}$$u - \dfrac{2}{u} = 1$
   2. Step 2: Multiply by u: a quadratic in u$u^{2} - u - 2 = 0$
   3. Step 3: Solve for u$(u + 1)(u - 2) = 0$$u = -1$ or $u = 2$
   4. Step 4: Back to x: lg x = u means x = 10ᵘ$x = \dfrac{1}{10}$ or $x = 100$