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1) If $\left(3^{x+1}\right)\left(4^{x+2}\right)=5^{x+3}$ , find the values of $x\,$ correct to two significant figures.

2) Without using mathematical tables or a calculator, solve the equation $4^{\left(2x-1\right)^{2}}=2^{x+1}$

3) If $1+\log_{a}\left(7x-3a\right)=2\log_{a}x+\log_{a}2$ , find in terms of $a\,$ , the possible values of $x\,$ .

4) Solve the equation $x-5y=0, \left(1-\log_{5}y\right)\log_{x}5=1$

5) Solve the equation $\log_{3}x=\log_{x}9\,$ , giving your answers correct to three significant figures.

6) Solve the equation $7\left(3^{x}\right)=15-2\left(3^{-x}\right)$ . Give your answers correct to three significant figures.

7) Solve the simultaneous equations $\log_{4}\left(xy\right)=\frac{1}{2}$ and $\left(\log_{2}x\right)\left(\log_{2}y\right)=-2$

8) Find the non-zero value of $x\,$ that satisfies the equation $3^{2x}-3^{x+1}+2=0\,$ . Give your answers correct to three decimal place.

9) Solve the simultaneous equation $x^{\frac{1}{2}}+y^{\frac{2}{3}}=6$ , $x^{-\frac{1}{2}}+y^{-\frac{2}{3}}=\frac{3}{4}$

10) Solve the equation $\sqrt{x+3}-\sqrt{x-2}=\sqrt{x-5}$

11) Solve the equation $\sqrt{3x+1}+\sqrt{x}=3$ [The expression $\sqrt{u}$ represents the positive square root]

12) Express $\sqrt{59-24\sqrt{6}}$ as $p\sqrt{2}+q\sqrt{3}$ where $p\,$ and $q\,$ are integers.

Answers

1) If $\left(3^{x+1}\right)\left(4^{x+2}\right)=5^{x+3}$ , find the values of $x\,$ correct to two significant figures.

$x=1.1\,$
- $\begin{align} & \left(3^{x+1}\right)\left(4^{x+2}\right)=5^{x+3} \\ & 3\left(3^{x}\right)\cdot 16\left(4^{x}\right)=125\cdot 5^{x} \\ & \frac{3^{x}\cdot 4^{x}}{5^{x}}=\frac{125}{48}\\ & \left(\frac{12}{5}\right)^{x}=\frac{125}{48}\\ & \lg \left(\frac{12}{5}\right)^{x}=\lg \left(\frac{125}{48}\right)\\ & x \lg \left(\frac{12}{5}\right)=\lg \left(\frac{125}{48}\right)\\ & x =\frac{\lg \left(\frac{125}{48}\right)}{\lg \left(\frac{12}{5}\right)}\\ & x = 1.1 \mbox { (2 s.f.) } \end{align}$

2) Without using mathematical tables or a calculator, solve the equation $4^{\left(2x-1\right)^{2}}=2^{x+1}$

$x=\frac{1}{8} \mbox{ or } x=1 \,$
- $\begin{align} & 4^{\left(2x-1\right)^{2}}=2^{x+1} \\ & 2^{2\left(2x-1\right)^{2}}=2^{x+1} \\ & \therefore 2\left(2x-1\right)^{2} = x+1 \\ & 2\left(4x^{2}-4x+1\right)=x+1 \\ & 8x^{2}-8x+2=x+1 \\ & 8x^{2}-9x+1=0 \\ & \left(8x-1\right)\left(x-1\right)=0\\ & x=\frac{1}{8} \mbox{ or } x=1 \\ \end{align}$

3) If $1+\log_{a}\left(7x-3a\right)=2\log_{a}x+\log_{a}2$ , find in terms of $a\,$ , the possible values of $x\,$ .

$x=\frac{1}{2}a \mbox{ or } x=3a \,$
- $\begin{align} & 1+\log_{a}\left(7x-3a\right)=2\log_{a}x+\log_{a}2 \\ & 1=2\log_{a}x+\log_{a}2-\log_{a}\left(7x-3a\right)\\ & \log_{a}\left(\frac{2x^{2}}{7x-3a}\right)=1 \\ & \therefore \frac{2x^{2}}{7x-3a} =a \\ & 2x^{2}=a\left(7x-3a\right)\\ & 2x^{2}=7ax-3a^{2} \\ & 2x^{2}-7ax+3a^{2}=0 \\ &\left(2x-a\right)\left(x-3a\right)=0\\ & x=\frac{1}{2}a \mbox{ or } x=3a \\ \end{align}$

4) Solve the equation $x-5y=0, \left(1-\log_{5}y\right)\log_{x}5=1$

$x=5, y=1\,$
- $\begin{align} & x-5y=0 \qquad\qquad \left(1-\log_{5}y\right)\log_{x}5=1 \frac{\qquad}{}(2)\\ & x=5y\frac{\qquad}{}(1) \\ &(1)\to(2):\\ &\left(1-\log_{5}y\right)\log_{5y}5=1\\ &\left(1-\log_{5}y\right)\frac{1}{\log_{5}5y}=1\\ &1-\log_{5}y=\log_{5}5y\\ &1-\log_{5}y=1+\log_{5}y\\ &\log_{5}y=0\\ & \therefore y=1, x=5 \\ \end{align}$

5) Solve the equation $\log_{3}x=\log_{x}9\,$ , giving your answers correct to three significant figures.

$x =4.72 \mbox{ or } x =0.211\,$
- $\begin{align} &\log_{3}x=\log_{x}9 \\ &\log_{3}x=2\log_{x}3 \\ &\log_{3}x=\frac{2}{\log_{3}x} \\ &\left(\log_{3}x\right)^{2}=2 \\ &\log_{3}x=\pm \sqrt{2} \\ & x =3^{\sqrt{2}} \mbox{ or } x =3^{-\sqrt{2}} \\ & x =4.72 \mbox{ or } x =0.211 \mbox{ 3 s.f. }\\ \end{align}$

6) Solve the equation $7\left(3^{x}\right)=15-2\left(3^{-x}\right)$ . Give your answers correct to three significant figures.

$x= -1.77 \mbox{ or }x=0.631\,$
- $\begin{align} & 7\left(3^{x}\right)=15-2\left(3^{-x}\right) \\ & 7\left(3^{x}\right)=15-\frac{2}{3^{x}}\\ & \mbox{Let } u=3^{x} \\ & 7u=15-\frac{2}{u}\\ & 7u^{2}=15u-2\\ & 7u^{2}-15u+2=0\\ & \left(7u-1\right)\left(u-2\right)=0\\ & u=\frac{1}{7} \mbox{ or } u=2\\ & 3^{x}=\frac{1}{7} \mbox{ or } 3^{x}=2\\ & x=\frac{\lg\left(\frac{1}{7}\right)}{\lg 3} \mbox{ or } x=\frac{\lg 2}{\lg 3}\\ & x= -1.77 \mbox{ or }x=0.631 \end{align}$

7) Solve the simultaneous equations $\log_{4}\left(xy\right)=\frac{1}{2}$ and $\left(\log_{2}x\right)\left(\log_{2}y\right)=-2$

$x=4, y=\frac{1}{2} \mbox{ or } x=\frac{1}{2}, y=4$
- $\begin{align} &\log_{4}\left(xy\right)=\frac{1}{2} \qquad \qquad \left(\log_{2}x\right)\left(\log_{2}y\right)=-2\\ & \frac{\log_{2}\left(xy\right)}{\log_{2}4}=\frac{1}{2}\\ & \frac{\log_{2}x+\log_{2}y}{2}=\frac{1}{2}\\ & \log_{2}x+\log_{2}y=1\\ & \mbox{Let }a=\log_{2}x, b=\log_{2}y\\ & a+b=1 \qquad \qquad \qquad ab=-2\frac{\qquad}{}(2) \\ & b=1-a\frac{\qquad}{}(1) \\ & (1)\to(2):\\ & a\left(1-a\right)=-2 \\ & a-a^{2}=-2 \\ & a^{2}-a-2=0 \\ & \left(a-2\right)\left(a+1\right)=0\\ & a=2 \mbox{ or } a=-1 \\ & \mbox{When } a=2, b=-1 \\ & \mbox{When } a=-1, b=2 \\ & \therefore \log_{2}x=2,\log_{2}y=-1\\ & x=4, y=\frac{1}{2}\\ & \mbox{or } \log_{2}x=-1,\log_{2}y=2\\ & x=\frac{1}{2}, y=4\\ & \therefore x=4, y=\frac{1}{2} \mbox{ or } x=\frac{1}{2}, y=4\\ \end{align}$

8) Find the non-zero value of $x\,$ that satisfies the equation $3^{2x}-3^{x+1}+2=0\,$ . Give your answers correct to three decimal place.

$x= 0.631\,$
- $\begin{align} & 3^{2x}-3^{x+1}+2=0\\ & \left(3^{x}\right)^{2}-3\left(3^{x}\right)+2=0\\ & \mbox{Let } u=3^{x}\\ & u^{2}-3u+2=0\\ & \left(u-2\right)\left(u-1\right)=0\\ & u=2 \mbox{ or } u=1 \\ & 3^{x}=2 \mbox{ or } 3^{x}=1 \\ & x=\frac{\lg 2}{\lg 3} \mbox{ or } x=0 \mbox{ (rejected)}\\ & x= 0.631 \mbox{ (3 d.p.)} \end{align}$

9) Solve the simultaneous equation $x^{\frac{1}{2}}+y^{\frac{2}{3}}=6$ and $x^{-\frac{1}{2}}+y^{-\frac{2}{3}}=\frac{3}{4}$

$x=16, y=\sqrt{8} \mbox{ or } x=4, y=8\,$
- $\begin{align} & x^{\frac{1}{2}}+y^{\frac{2}{3}}=6 \qquad \qquad x^{-\frac{1}{2}}+y^{-\frac{2}{3}}=\frac{3}{4}\\ & \qquad \qquad \qquad \qquad \quad \frac{1}{x^{\frac{1}{2}}}+\frac{1}{y^{\frac{2}{3}}}=\frac{3}{4}\\ & \mbox{Let } a=x^{\frac{1}{2}}, b=y^{\frac{2}{3}} \\ & a+b=6 \qquad \qquad \frac{1}{a}+\frac{1}{b}=\frac{3}{4}\frac{\qquad}{}(2)\\ & b=6-a\frac{\qquad}{}(1) \\ & (1)\to(2):\\ & \frac{1}{a}+\frac{1}{6-a}=\frac{3}{4}\\ & \frac{\left(6-a\right)+a}{a\left(6-a\right)}=\frac{3}{4}\\ & \frac{6}{a\left(6-a\right)}=\frac{3}{4}\\ & 8=a\left(6-a\right)\\ & 8=6a-a^{2}\\ & a^{2}-6a+8=0\\ & \left(a-4\right)\left(a-2\right)=0\\ & a=4 \mbox{ or } a=2 \\ & \mbox{When } a=4, b=2 \\ & \mbox{When } a=2, b=4 \\ & \therefore x^{\frac{1}{2}}=4, y^{\frac{2}{3}}=2\\ & x=16, y=2^{\frac{3}{2}}=\sqrt{8} \\ & \mbox{or } x^{\frac{1}{2}}=2, y^{\frac{2}{3}}=4\\ & x=4, y=4^{\frac{3}{2}}=8 \\ & \therefore x=16, y=\sqrt{8} \mbox{ or } x=4, y=8\, \\ \end{align}$

10) Solve the equation $\sqrt{x+3}-\sqrt{x-2}=\sqrt{x-5}$

$x=6\,$
- $\begin{align} & \sqrt{x+3}-\sqrt{x-2}=\sqrt{x-5}\\ & \left(\sqrt{x+3}-\sqrt{x-2}\right)^{2}=\left(\sqrt{x-5}\right)^{2}\\ & \left(x+3\right)-2\sqrt{x+3}\sqrt{x-2}+\left(x-2\right)=x-5\\ & x+6=2\sqrt{x+3}\sqrt{x-2}\\ & x+6=2\sqrt{\left(x+3\right)\left(x-2\right)}\\ & \left(x+6\right)^{2}=4\left(x+3\right)\left(x-2\right)\\ & x^{2}+12x+36=4\left(x^{2}+x-6\right)\\ & x^{2}+12x+36=4x^{2}+4x-24\\ & 3x^{2}-8x-60=0\\ & \left(3x-10\right)\left(x-6\right)=0\\ & x=-\frac{10}{3}\mbox{ (rejected) } \mbox{ or } x=6 \\ & \mbox{When } x=6, \mbox{R.H.S }=\sqrt{9}-\sqrt{4}=3-2=1, \mbox{L.H.S }=\sqrt{1}=1=\mbox{R.H.S}\\ & \therefore x=6 \\ \end{align}$

11) Solve the equation $\sqrt{3x+1}+\sqrt{x}=3$ [The expression $\sqrt{u}$ represents the positive square root]

$x=1\,$
- $\begin{align} & \sqrt{3x+1}+\sqrt{x}=3 \\ & \sqrt{3x+1}=3-\sqrt{x} \\ & 3x+1=\left(3-\sqrt{x}\right)^{2} \\ & 3x+1=9-6\sqrt{x}+x\\ & 6\sqrt{x}=8-2x\\ & 3\sqrt{x}=4-x\\ & 9x=\left(4-x\right)^{2}\\ & 9x=16-8x+x^{2}\\ & x^{2}-17x+16=0\\ & \left(x-16\right)\left(x-1\right)=0\\ & x=16\mbox{ (rejected) } \mbox{ or } x=1 \\ & \mbox{When } x=16, \mbox{R.H.S }=\sqrt{49}+\sqrt{16}=7+4=11\neq\mbox{R.H.S }\\ & \mbox{When } x=1, \mbox{R.H.S }=\sqrt{4}+\sqrt{1}=2+1=3=\mbox{R.H.S }\\ & \therefore x=1\\ \end{align}$

12) Express $\sqrt{59-24\sqrt{6}}$ as $p\sqrt{2}+q\sqrt{3}$ where $p\,$ and $q\,$ are integers.

$\sqrt{59-24\sqrt{6}}=4\sqrt{2}-3\sqrt{3}$
- $\begin{align} &\mbox{Let } \sqrt{59-24\sqrt{6}}=p\sqrt{2}+q\sqrt{3}\\ &\therefore 59-24\sqrt{6}=\left(p\sqrt{2}+q\sqrt{3}\right)^{2}\\ & 59-24\sqrt{6}=2p^{2}+2\left(p\sqrt{2}\right)\left(q\sqrt{3}\right)+3q^{2}\\ & 59-24\sqrt{6}=2p^{2}+3q^{2}+2pq\sqrt{6}\\ &\therefore 59=2p^{2}+3q^{2}\frac{\qquad}{}(1) \qquad\qquad -24=2pq \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad q=-\frac{12}{p}\frac{\qquad}{}(2)\\ &(2)\to(1):\\ & 59=2p^{2}+3\left(-\frac{12}{p}\right)^{2} \\ & 59=2p^{2}+\frac{432}{p^{2}} \\ & 59p^{2}=2p^{4}+432 \\ & 2p^{4}-59p^{2}+432=0 \\ & \left(2p^{2}-17\right)\left(p^{2}-16\right)=0\\ & p^{2}=\frac{17}{2}\mbox{ (rejected) } \mbox{ or } p^{2}=16\\ & \qquad \qquad \qquad \qquad \qquad p=\pm 4\\ & \mbox{When } p=4, q=-3 , \sqrt{59-24\sqrt{6}}=4\sqrt{2}-3\sqrt{3}>0 \\ & \mbox{When } p=-4, q=3 , \sqrt{59-24\sqrt{6}}=-4\sqrt{2}+\sqrt{3}<0 \mbox{ (rejected) }\\ & \therefore \sqrt{59-24\sqrt{6}}=4\sqrt{2}-3\sqrt{3} \\ \end{align}$

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