Logarithm Functions — Question 1

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Question 1

Solve the equation for xx, showing all steps: log⁡3(x+5)+log⁡3(x−2)=2\log_3(x + 5) + \log_3(x - 2) = 2

  • (a) Solve the logarithmic equation algebraically.

  • (b) State any restrictions on the domain.

  • (c) Check for extraneous solutions.

Original worksheet page 1: question and worked solution for 1-8-001
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Question 1 - Solution

(a) Combine the logs using log rules: log⁡3(x+5)+log⁡3(x−2)=log⁡3[(x+5)(x−2)]=2⇒log⁡3(x2+3x−10)=2\log_3(x + 5) + \log_3(x - 2) = \log_3[(x + 5)(x - 2)] = 2 \Rightarrow \log_3(x^2 + 3x - 10) = 2

Rewrite in exponential form: x2+3x−10=32=9⇒x2+3x−10=9⇒x2+3x−19=0x^2 + 3x - 10 = 3^2 = 9 \Rightarrow x^2 + 3x - 10 = 9 \Rightarrow x^2 + 3x - 19 = 0

Solve with quadratic formula: x=−3±32−4(1)(−19)2(1)=−3±9+762=−3±852x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-19)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 76}}{2} = \frac{-3 \pm \sqrt{85}}{2}

(b) Domain restrictions:

Since log⁡3(x+5)\log_3(x + 5) and log⁡3(x−2)\log_3(x - 2) are defined only if: x+5>0⇒x>−5,x−2>0⇒x>2⇒Overall domain: x>2x + 5 > 0 \Rightarrow x > -5,\quad x - 2 > 0 \Rightarrow x > 2 \Rightarrow \text{Overall domain: } x > 2

(c) Check solutions:

85≈9.22⇒x1=−3+9.222≈3.11,x2=−3−9.222≈−6.11\sqrt{85} \approx 9.22 \Rightarrow x_1 = \frac{-3 + 9.22}{2} \approx 3.11,\quad x_2 = \frac{-3 - 9.22}{2} \approx -6.11

Only x1≈3.11x_1 \approx 3.11 satisfies x>2x > 2. So:

x=−3+852(valid solution)\boxed{x = \frac{-3 + \sqrt{85}}{2}} \quad \text{(valid solution)}

Original worksheet page 2: question and worked solution for 1-8-001

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