Solve \(2 \log _3(x-4)+\log _3(x)=2\) for \(x\). (4 marks) --- 8 WORK AREA LINES (style=lined) ---
\(1^3-8(1)^2+16(1)-9=0\) \(\therefore\ x-1\ \text{is a factor} \) \((x-1)(x^2-7x+9)=0\) \(\text{Using quadratic formula to solve}\ \ x^2-7x+9=0:\) \( x=1, \dfrac{7- \sqrt{13}}{2}, \dfrac{7 + \sqrt{13}}{2}\) \(\therefore\ \dfrac{7 + \sqrt{13}}{2}\ \text{ is the only possible solution.}\)
\(2\log_3(x-4)+\log_3(x)\)
\(=2\)
\(\log_3x(x-4)^2\)
\(=2\)
\(x(x-4)^2\)
\(=3^2\)
\(x(x^2-8x+16)-9\)
\(=0\)
\(x^3-8x^2+16x-9\)
\(=0\)
\(\text{Find a factor}\ \ \Rightarrow\ \ \text{Test}\ \ x=1:\)
\(x\)
\(=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(1)(9)}}{2(1)}\)
\(=\dfrac{7\pm \sqrt{49-36}}{2}\)
\(=\dfrac{7\pm \sqrt{13}}{2}\)
\(\text{For }\log_3(x-4)\ \text{to exist}\ x>4\)
Algebra, MET1 SM-Bank 8
Write `log2 + log4 + log8 + log16 + … + log128` in the form `a logb` where `a` and `b` are integers greater than 1. (2 marks)
Algebra, MET2 SM-Bank 1 MC
Let `a = e^x`
Which expression is equal to `log_e(a^2)`?
- `e^(2x)`
- `e^(x^2)`
- `2x`
- `x^2`
- `e^2`
Algebra, MET2 2010 VCAA 8 MC
The function `f` has rule `f(x) = 3 log_e (2x).`
If `f(5x) = log_e (y)` then `y` is equal to
- `30x`
- `6x`
- `125x^3`
- `50x^3`
- `1000x^3`