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Given that $$2 ext{log}_2{(x+15)} - ext{log}_2{x} = 6$$ (a) Show that $$x^2 - 34x + 225 = 0$$ (b) Hence, or otherwise, solve the equation $$2 ext{log}_2{(x+15)} - ext{log}_2{x} = 6$$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 6
Question 7
Given that $$2 ext{log}_2{(x+15)} - ext{log}_2{x} = 6$$ (a) Show that $$x^2 - 34x + 225 = 0$$ (b) Hence, or otherwise, solve the equation $$2 ext{log}_2{(x+15)... show full transcript
Worked Solution & Example Answer:Given that $$2 ext{log}_2{(x+15)} - ext{log}_2{x} = 6$$ (a) Show that $$x^2 - 34x + 225 = 0$$ (b) Hence, or otherwise, solve the equation $$2 ext{log}_2{(x+15)} - ext{log}_2{x} = 6$$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 6
Step 1
Show that $$x^2 - 34x + 225 = 0$$
Answer
To derive the quadratic equation, we start with the given logarithmic equation:
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Rewrite the equation using properties of logarithms.
From the logarithmic property, we have:Thus, the equation becomes:
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Combine the logarithms.
Using the law of logarithms, we get:ext{log}_2{ rac{(x+15)^2}{x}} = 6
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Convert the logarithmic statement to exponential form.
This gives us:rac{(x+15)^2}{x} = 2^6
Simplifying, we find:
rac{(x+15)^2}{x} = 64
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Multiply through by x to eliminate the fraction.
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Expand the left side.
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Rearrange the equation.
Step 2
Hence, or otherwise, solve the equation $$2 ext{log}_2{(x+15)} - ext{log}_2{x} = 6$$
Answer
Having established the quadratic equation:
we can solve for x using the quadratic formula:
ightarrow ext{coefficients}}{2a}$$ where \(a = 1\) and \(b = -34\) and \(c = 225\). 1. **Substituting the values into the formula:** $$x = \frac{34 \pm \sqrt{(-34)^2 - 4(1)(225)}}{2(1)}$$ 2. **Calculate the discriminant:** $$\sqrt{1156 - 900} = \sqrt{256} = 16$$ 3. **Substituting back gives:** $$x = \frac{34 \pm 16}{2}$$ 4. **This results in two possible solutions:** - $$x = \frac{50}{2} = 25$$ - $$x = \frac{18}{2} = 9$$ Hence, the solutions are **25** and **9**.