Numerical concepts Past Questions & Worked Solutions

(i) $2^{2x+1}-7(2^x)+6=0$

$2^{2x}\cdot 2^1-7(2^x)+6=0$

$2^1(2^x)^2-7(2^x)+6=0$

Let $p=2^x$

$\Rightarrow\quad 2p^2-7p+6=0$

$2p^2-4p-3p+6=0$

$2p(p-2)-3(p-2)=0$

$(2p-3)(p-2)=0$

Either $2p-3=0$

$p=\frac{3}{2}$

Or $p-2=0$

$p=2$

When $p=\dfrac{3}{2}\ \Rightarrow\ 2^x=\dfrac{3}{2}$

$\log 2^x=\log\left(\frac{3}{2}\right)$

$x\log 2=\log\left(\frac{3}{2}\right)$

$\therefore\ x=\frac{\log\left(\frac{3}{2}\right)}{\log 2}=0.5850$

When $p=2\ \Rightarrow\ 2^x=2=2^1$

$x=1$

Hence $x=1$ and $x=0.585$ (3.d.p)

(ii) $3^{2x+1}+3=10\,(3^x)$

$3^{2x}\cdot 3+3=10\cdot 3^x$

$3(3^x)^2-10(3^x)+3=0$

Let $m=3^x$

$3m^2-10m+3=0$

$3m^2-9m-m+3=0$

$3m(m-3)-1(m-3)=0$

$(3m-1)(m-3)=0$

Either $3m-1=0\ \Rightarrow\ m=\dfrac{1}{3}$

Or $m-3=0\ \Rightarrow\ m=3$

When $m=\dfrac{1}{3}\ \Rightarrow 3^x=\dfrac{1}{3}=3^{-1}$

By equating powers;

$x=-1$

When $m=3\ \Rightarrow 3^x=3^1$

By equating powers; $x=1$

Hence $x=1$ and $x=-1$

(iii) $2^{2x-1}+\dfrac{3}{2}=2^{x+1}$

Multiplying through by $2$;

$2\cdot 2^{2x}\cdot 2^{-1}+3=2\cdot 2^x\cdot 2^1$

$(2^x)^2-4(2^x)+3=0$

Let $p=2^x$

$p^2-4p+3=0$

$p^2-p-3p+3=0$

$p(p-1)-3(p-1)=0$

$(p-1)(p-3)=0$

Either $p-1=0\ \Rightarrow p=1$

Or $\quad p-3=0\ \Rightarrow p=3$

When $p=1\ \Rightarrow 2^x=1=2^0$

$x=0$

When $p=3\ \Rightarrow 2^x=3$

Introducing log to base ten on both sides;

$\log 2^x=\log 3$

$x\log 2=\log 3$

$x=\dfrac{\log 3}{\log 2}=1.585$

Hence $x=0$ and $x=1.585$

(iv) $7^x=3^{1-x}$

$\log 7^x=\log 3^{1-x}$

$x\log 7=(1-x)\log 3$

$\log 3=x(\log 3+\log 7)$

$x=\frac{\log 3}{\log 3+\log 7}=\frac{\log 3}{\log 21}=0.3608$

$$\log_2 x+2\log_4 y=4$$ $$\log_2 x+\log_4 y^2=4$$ $$\log_2 x+\frac{\log_2 y^2}{\log_2 2^2}=4$$ $$\log_2 x+\frac{\log_2 y^2}{2}=4$$ $$2\log_2 x+\log_2 y^2=8$$ $$\log_2 x^2+\log_2 y^2=8$$ $$\log_2 x^2y^2=8$$ $$x^2y^2=2^8$$ $$xy=2^4=16$$

Solving equations;

$$\log_{10}(x+y)=1$$ $$x+y=10^1=10$$ $$x=10-y \tag{i}$$ $$\log_2 x+2\log_4 y=4$$ $$xy=16 \tag{ii}$$

Substituting eqn (i) into eqn (ii)

$$(10-y)y=16$$ $$10y-y^2=16$$ $$y^2-10y+16=0$$$$y^2-8y-2y+16=0$$ $$(y-8)(y-2)=0$$

Either: $y-8=0$

$$y=8$$

Or: $y-2=0$

$$y=2$$

If $y=8$; $x=10-8=2$

If $y=2$; $x=10-2=8$

Hence $(x,y)=(2,8)$ or $(8,2)$