Solve \(2 \log _3(x-4)+\log _3(x)=2\) for \(x\). (4 marks) --- 8 WORK AREA LINES (style=lined) ---
\(1^3-8(1)^2+16(1)-9=0\) \(\therefore\ x-1\ \text{is a factor} \) \((x-1)(x^2-7x+9)=0\) \(\text{Using quadratic formula to solve}\ \ x^2-7x+9=0:\) \( x=1, \dfrac{7- \sqrt{13}}{2}, \dfrac{7 + \sqrt{13}}{2}\) \(\therefore\ \dfrac{7 + \sqrt{13}}{2}\ \text{ is the only possible solution.}\)
\(2\log_3(x-4)+\log_3(x)\)
\(=2\)
\(\log_3x(x-4)^2\)
\(=2\)
\(x(x-4)^2\)
\(=3^2\)
\(x(x^2-8x+16)-9\)
\(=0\)
\(x^3-8x^2+16x-9\)
\(=0\)
\(\text{Find a factor}\ \ \Rightarrow\ \ \text{Test}\ \ x=1:\)
\(x\)
\(=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(1)(9)}}{2(1)}\)
\(=\dfrac{7\pm \sqrt{49-36}}{2}\)
\(=\dfrac{7\pm \sqrt{13}}{2}\)
\(\text{For }\log_3(x-4)\ \text{to exist}\ x>4\)
L&E, 2ADV E1 2020 MET1 4
Solve the equation `2 log_2(x + 5)-log_2(x + 9) = 1`. (3 marks)
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