Exercise Set 3 Answers

  1. Solve the following quadratic equations by factorisation.

    1. \(x^2-28x-60=0\)
    2. \(p^2=8p-15\)
    3. \(3x^2-14x+8\)
    4. \(x^2+10x+25\)

    Answers:

    1. We are looking for two factors of 25 with summation equal to 10. Thus, these factors are 5 and 5

    \[ x^2 + 10x + 25 = (x+5)(x+5) = (x+5)^2 \] So there is a single solution \(x=-5\).

    1. Factorise by inspection into the form \((x+)(x+b)\) where \(ab = -60\) and \(a + b = -28\)

    \[\begin{align*} x^2-28x-60 &= 0\\ (x-30)(x+2) & = 0\\ \Rightarrow x=30, -2 \end{align*}\]

    1. First, put into quadratic form \(ax^2 + bx + c = 0\) and then factorise by inspection

    \[\begin{align*} p^2 = 8p - 18\\ p^2 - 8p + 15 &= 0\\ (p-3)(p-5) &= 0\\ \Rightarrow p = 3,5 \end{align*}\]

    1. We are looking for two factors of \(3\times 8 = 24\) with summation equals to -14. These factors are \(-12\) and \(-2\) \[\begin{align*} 3x^2 - 14x + 8 &= 3x^2 - 12x - 2x + 8\\ &= 3x(x-4) - 2(x-4) = (x-4)(3x-2) \end{align*}\]

  2. Solve the following quadratic equations, giving results correct to 2 d.p.

    1. \(4x^2+x-3=0\)
    2. \(x^2+x=5\)
    3. \(x+\frac{1}{x}=5\)

    Answers:

    These questions require the application of the quadratic formula: for a quadratic which can be put into the form \(ax^2 + bx + c = 0\), \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

    1. \(4x^2 + x - 3\)

    \[ x = \frac{-1 \pm \sqrt{1^2 - 4\times 4 \times (-3)}}{8}\\ x = 3.38, -3.63 \]

    1. First, rearrange \(x^2 + x = 5\) as \(x^2 + x - 5 =0\), then apply the quadratic formula \[ x = \frac{-1 \pm \sqrt{1^2 - 4\times 1 \times (-5)}}{2}\\ x = 1.79, -2.79 \]
    2. First, rearrange \(x + \frac{1}{x} = 5\) into \(x^2 + 1 - 5 = 0\) and then apply the quadratic formula. \[ x = \frac{5 \pm \sqrt{(-5)^2 - 4\times 1 \times 1}}{2}\\ x = 4.79, 0.21 \]

  3. Solve the following sets of simultaneous equations.

    1. \(3x+4y=7\), \(5x+6y=11\)
    2. \(2x+y=7\), \(x^2-xy=6\)
    3. \(x + y=2\), \(x^2-xy+y^2 = 1\)

    Answers:

    1. Because these simultaneous equations are linear in \(x\) and \(y\), they can be solved by multiplying the equations and adding or subtracting them to cancel one of \(x\) or \(y\). Here, they can be solved by multiplying the left equation by 5 and the right equation by 3, then subtracting to cancel the \(x\) term: \[ 3x+4y = 7,\quad 5x + 6y = 11\\ 15 x + 20y = 35,\quad 15x + 18y = 33\\ \text{subtracting the right equaltion from the left:}\\ 2y = 2 \Rightarrow y=1,x=1 \] This could also have been solved via direct substitution (e.g., \(x = (7-4y)/3\)).

    2. \[ 2x + y = 7,\quad x^2 - xy = 6\\ \Rightarrow y=7-2x \. \text{substitute into right hand equation}\\ x^2 - x(7-2x) = 6\\ x^2 - 7x + 2x^2 = 6\\ 3x^2 - 7x - 6 = 0 \] which can now be factorised by inspection to obtain the two solutions for \(x\) \[ (3x+2)(x-3) = 0\\ \Rightarrow x = -\frac{2}{3},x = 3 \] The value for \(y\) must be calculated for both cases of \(x\). It is simplest to use \(y = 7 - 2x\). The solutions are: \[ x = -\frac{2}{3},y=\frac{25}{3},\quad\text{and }\, x = 3, y = 1 \]

    3. As before, substitute one of \(x\) or \(y\) using the linear equation \[ x+y = 2, \quad x^2 - xy + y^2 = 1\\ \Rightarrow x = 2-y,\quad (2-y)^2 - (2-y)y + y^2 = 1\\ 4 - 4y^2 + y^2 - 2y + y^2 + y^2 = 1\\ y^2 + 2y - 3 = 0\\ (y+3)(y-1) = 0\\ y=1, -3 \] and again, we must give \(x\) for both cases. The two solutions are \[ x=1,y=1\quad\text{and }x =5,y=-3 \]

  4. Show that the following hyperbolic identity holds: \[\cosh^2(x)-\sinh^2(x)=1.\]

Answer: Use the definitions of \(\cosh(x) = \frac{e^x + e^-x}{2}\) and \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) \[ \begin{align} \cosh^2(x) - \sinh^2(x) &=(\cosh(x) + \sinh{x})(\cosh(x) - \sinh(x))\\ &=\left(\frac{e^x + e^{-x} + e^x - e^{-x}}{2}\right)\left(\frac{e^x + e^{-x} - e^x + e^{-x}}{2}\right)\\ &=e^x e^{-x}\\ &= 1 \text{ as required} \end{align} \]

  1. Use natural logarthims to make \(t\) the subject of the formula \[V=1-e^{\frac{-t}{RC}}.\]

Answer: Make the exponential term the subject and then take the natural logarithms of both sides \[ \begin{align} V &= 1- e^{\frac{-t}{RC}}\\ e^{\frac{-t}{RC}} &= 1- v\\ \ln\left(e^{\frac{-t}{RC}}\right) &= \ln(1- v)\\ \frac{-t}{RC} &= \ln(1-v)\\ t &= -RC\ln(1-v) \end{align} \]

  1. Find the solutions to the following equations without using a calculator.
    1. \(x=\log_4(64)\)
    2. \(x=\log_{101}(101)\)
    3. \(x=\ln(e)\)
    4. \(x=\log_{56}(1)\)
    5. \(10^x=100000\)
    6. \(2^x=128\)
    7. \(\log_3(x)=4\)
    8. \(\log_x(125)=3\)
    Answers:
    1. \(x = \log_4(64) = \log_4(4^3)=3\)
    2. \(x = \log_{101}(101) =\log_{101}(101^1)=1\)
    3. \(x = \ln(e) =\log_e(e^1)= 1\)
    4. \(x = log_{56}(1) =\log_{56}(56^0)=0\)
    5. \(10^x = 100000=10^5 \Rightarrow x = 5\)
    6. \(2^x = 128=2^7\) x = 7$ the powers of 2 are useful to know
    7. \(3^{\log_3(x)} = 3^4 \Rightarrow x = 3^4 = 3^2 \times 3^2 = 81\)
    8. \(\log_x(125) = 3 \Rightarrow x^3 = 125 \Rightarrow x = 5\)
  2. Calculate the following logarithms using a calculator.
    1. \(\ln(2)\)
    2. \(\log(20)\)
    3. \(\log_{16}(100)\) - do this using your \(\log_a(x)\) button and also using the change of base rule with your \(\log\) or \(\ln\) button.
    Answers: To 2 d.p.
    1. \(\ln(2) = 0.69\)
    2. \(\log(20) = 1.30\)
    3. \(\log_{16}(100) = 1.66\) or using the change of base rule with \(\log_{10}\), \(\log_{16}=\frac{\log_{10}(100)}{\log_{10}(16)}=\frac{2}{\log_{10}(16)}=1.66\)
  3. Calculate \(4321\times 9876\) using logs: convert both values to logs, add, then convert back.

Answer: \[\log(4321 \times 9876)=\log(4321) + \log(9876) = 7.630165..., \text{ the product rule of logarithms}\\ 4321 \times 9876=10^{7.630165...} = 42674196 \]

  1. Simplify the following expression involving logarithms. \[\log_a(x^2)+3\log_a(x)-2\log_a(4x^2).\]

Answer: Applying the various log rules, starting with the power rules then product and quotient rules: \[ \begin{aligned} & \log_a(x^2)+3\log_a(x)-2\log_a(4x^2)\\ &= \log_a(x^2) + \log_a(x^3) - log_a(16x^4)\\ &= \log_a(\frac{x^5}{16x^4})\\ &= \log_a(\frac{x}{16}) \end{aligned} \]

  1. Solve the following for \(x\). \[2\log_a(x)-\log_a(x-1)=\log_a(x+3).\]

Answer: \[ \begin{aligned} 2 \log_a(x) - \log_a(x-1) &= \log_a(x+3)\\ \log_a\left(\frac{x^2}{x-1}\right) &= \log_a(x+3)\\ \log_a\left(\frac{x^2}{(x-1)(x+3)}\right) &= 0, \text{ since } \log_a(1) = 0\\ \frac{x^2}{(x-1)(x+3)} &= 1\\ x^2 &= (x-1)(x+3) = x^2 + 2x - 3\\ x &= \frac{3}{2} \end{aligned} \]

  1. Solve the following for \(x\). \[\log_a(x^2-10)-\log_a(x)=2\log_a(3).\]

Answer: \[ \begin{align} \log_a(x^2-10)-\log_a(x)&=2\log_a(3)\\ \log_a(x^2-10) &= \log_a(x) + \log_a(3^2)\\ \log_a(x^2-10) &= \log_a(9x)\\ x^2-10&=9x\\ x^2-9x-10&=0\\ (x+1)(x-10)&=0\\ \end{align} \] Hence \(x=-1\) or \(x=10\).

  1. Bacteria are undergoing cell division every 30 minutes. Approximating the number of bacteria as a continuous variable \(x\), if there are initially 5 bacteria, write an equation for the number of bacteria \(x\) at time \(t\) in the form \[x=A2^{kt}\] where \(t\) is measured in minutes. Also express this in the form \[x=Ae^{\lambda t}\] by finding the appropriate value of \(\lambda\).

Answers: Since \(t\) is in minutes and the bacteria double every 30 mins, we have \[x=A2^{\frac{1}{30}t}.\] At the initial time \(t=0\) \[x=A2^{\frac{1}{30}\times 0}=A2^0=A\times 1=A\] hence \(A=5\) and we have \[x=5\times2^{\frac{1}{30}t}.\]

We can write \(2^{\frac{1}{30}t}=(2^{\frac{1}{30}})^t\) and hence \(\lambda=\ln(2^{\frac{1}{30}})=\frac{\ln(2)}{30}\). Then \[x=5e^{\frac{\ln(2)}{30}t}.\]

  1. A quantity \(x\) is increasing exponentially with respect to time \(t\). We have the measurements \(x=21\) at \(t=0\) and \(x=156\) at \(t=10\). Find an equation for \(x\) in the form \(x=Ae^{kt}\).

Answers: At \(t=0\) \[x=Ae^{k\times0}=A\times 1=21\] so \(A=21\).

At \(t=10\) \[x=21e^{k\times 10}=156.\] Taking natural logs \[\ln(21e^{10k})=\ln(21)+10k=\ln(156)\] hence \[k=\frac{1}{10}\left(\ln(\frac{156}{21})\right)=0.20 (2 d.p.)\]. Then \[x=21e^{0.2t}.\]

  1. The number of specimens of an invasive plant species is observed to be increasing at a particular location. The number of plants observed last year was \(N=52\) and this year is \(N=76\). We shall approximate the number of plants as a continuous variable \(x\).

    1. Assuming continuous exponential growth \(x=Ae^{\lambda t}\), find the growth rate \(\lambda\).
    2. Approxiately how many plants are predicted by this model after 10 years from now?

Answer:

  1. We can take last year as the starting time \(t=0\), so that \(x=Ae^{\lambda\times 0}=A=52\). Then taking time in units of years, we have \[x=52e^{\lambda\times 1}=76.\] Taking natural logs of both sides, \[\ln(52)+\lambda=\ln(76)\] Hence \[\lambda = \ln\left(\frac{76}{52}\right)=0.38 (2 d.p.)\]

  2. 10 years from now corresponds to \(t=11\), so we have \[x=52e^{0.38\times 11}\approx 3380 \text{ plants}.\]

  1. Solve the following inequalities.

    1. \(7-3x>x-5\)
    2. \(x^2\ge 4\)
    3. \(x^2-x-6<0\)
    4. \(x^2-3x-12\le 2x+2\)
    5. \(\frac{2}{2x-1} > \frac{3}{3x+1}\)
    6. \(\frac{2x}{x-5}\le 3\)
    7. \(x^3-9x^2<0\)

    Answers:

    1. \[\begin{align*} 7-3x &> x-5\\ 13 &> 4x\\ x &< \frac{13}{4} \end{align*}\]

    2. Here, we must include both the positive and the negative solution to the square root: \[\begin{align*} x^2 &\geq 4\\ \pm x &\geq 2\\ x\geq 2 & x\leq -2 \end{align*}\] It is often worth sketching the graph and examining the inequality to determine where the inequality is satisfied.

    3. \[ x^2 - x - 6 < 0 \\ (x+2)(x-3) < 0 \] The graph of \(x^2\) will form a \(\cup\) shape, so the parts below the axis are between the crossing points of \(x = -2\) and \(x=3\). So the inequality is satisfied for \[ -2<x<3 \] Note the use of strict inequalities \(x<3\) and \(x>-2\) because the inequality is \(<0\) (as opposed to \(\leq\)).

    4. Group terms to form a quadratic, find the solutions and determine whether the interval between them or outside of them satisfied the inequality: \[\begin{align*} x ^2 - 3x - 12 &\leq 2x + 2\\ x^2 - 5x - 14 &\leq 0\\ (x-7)(x+2)&\leq 0 \end{align*}\] And the part of the curve below the axis is the range \(-2\leq x \leq 7\)

  1. Do not multiply through by a variable, since at times the term will be negative and necessitate a flip of the inequality. Instead, subtract one side from both sides:

    \[\begin{align*} \frac{2}{2x-1} &> \frac{3}{3x+1}\\ \frac{2}{2x-2} - \frac{3}{3x+1} &> 0\\ \frac{2(3x+1) - 3(2x-1)}{(2x-1)(3x+1)} &> 0\\ \frac{5}{(2x-1)(3x+1)} &> 0 \end{align*}\]

    In order for the inequality to hold, either both terms in the denominator are positive or both are negative. \[ \text{positve case: }2x-1>0,\quad 3x+1 > 0\\ x > \frac{1}{2} \text{ and } x > \frac{1}{3}\\ x > \frac{1}{2}\\ \text{negative case: }2x-2 < 0, \quad 3x+1 < 0\\ x<\frac{1}{2} \text{ and } x < \frac{1}{3}\\ x < \frac{1}{3} \] So the final solution is \(x > \frac{1}{2}\) and \(x < \frac{1}{3}\).

    1. \[\begin{align*} \frac{2x}{x-5} &\leq 3 \\ \frac{2x}{x-5} - 3 &\leq 0\\ \frac{2x - 3(x-5)}{x-5} & \leq 0\\ \frac{15-x}{x-5}&\leq 0 \end{align*}\]

    Which will ony old if one of the numerator or denominator is negative. So the inequality is solved by \(x\geq 15\) and \(x\leq 5\).

    1. Here, a cubic is presented, but a factor of \(x^2\) can be taken out \[\begin{align*} x^3 - 9x^2 &< 0 \\ x^2(x-9) &< 0\\ \end{align*}\] The \(x^2\) factor is always positive, so the whole inequality holds when \(x-9\) is negative. Hence, \(x<9\)