3 Mark Question

Question 13 Marks

Simplify:
$\frac{3^5\times10^5\times25}{5^7\times6^5}$

Answer

$\frac{3^5\times10^5\times25}{5^7\times6^5}=\frac{3^5\times(2\times5)^5\times5^2}{5^7\times(2\times3)^5}$
$=\frac{3^5\times2^5\times5^5\times5^2}{5^7\times2^5\times3^5}$
$=2^{5-5}\times3^{5-5}\times5^{5+2-7}$
$=2^0\times3^0\times5^0$
$=1\times1\times1$
$=1$

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Question 23 Marks

Find x such that $\Big(\frac{3}{5}\Big)^3\times\Big(\frac{3}{5}\Big)^{-6}=\Big(\frac{3}{5}\Big)^{2\text{x}-1}$

Answer

$\Big(\frac{3}{5}\Big)^3\times\Big(\frac{3}{5}\Big)^{-6}=\Big(\frac{3}{5}\Big)^{2\text{x}-1}$
$\Rightarrow \Big(\frac{3}{5}\Big)^{3-6}=\Big(\frac{3}{5}\Big)^{2\text{x}-1}$
$\Rightarrow \Big(\frac{3}{5}\Big)^{-3}=\Big(\frac{3}{5}\Big)^{2\text{x}-1}$
Comparing we get
$2\text{x}-1=-3$
$\Rightarrow 2\text{x}=-3+1$
$\Rightarrow 2\text{x}=-2$
$\Rightarrow \text{x}=\frac{-2}{2}=-1$
Hence x = -1

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Question 33 Marks

If $2^{\text{n}-7}\times5^{\text{n}-4}=1250,$ fimd the value of n.

Answer

We have,
$2^{n-7} \times 5^{n-4}=1250$
$\Rightarrow \frac{2^{n}}{2^7} \times \frac{5^{n}}{564}=2 \times 5^4\left[\text { Since } 1250=2 \times 5^4\right]$
$\Rightarrow \frac{2^{n} \times 5^{n}}{2^7 \times 5^4}=2 \times 5^4$
$\Rightarrow 2^{n} \times 5^{n}=2 \times 5^4 \times 2^7 \times 5^4 \text { [Using cross multiplication] }$
$\Rightarrow 2^{n} \times 5^{n}=2^{1+7} \times 5^{4+4}\left[\text { Since } a^{m} \times b^{n}=(a \times b)^{m+n}\right]$
$\Rightarrow 2^{n} \times 5^{n}=2^8 \times 5^8$
$\Rightarrow(2 \times 5)^{n}=(2 \times 5)^8\left[\text { Since } a^{n} \times b^{n}=(a \times b)^{n}\right]$
$\Rightarrow 10^{n}=10^8$
$\Rightarrow n=8$

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Question 43 Marks

Express the following in power notation:
$\frac{-32}{243}$

Answer

$\frac{-32}{243}=\frac{(-2)\times(-2)\times(-2)\times(-2)\times(-2) }{3\times3\times3\times3\times3}$
$\begin{array}{c|c} 2 & 32 \\ \hline 2 & 16\\ \hline2 &8\\ \hline 2 & 4\\ \hline2 & 2 \\ \hline &1 \end{array}$
$\begin{array}{c|c} 3 & 243 \\ \hline 3 & 81\\ \hline3&27\\ \hline3&9\\ \hline3&3\\ \hline&1 \end{array}$
$=\Big(\frac{-2}{3}\Big)\times \Big(\frac{-2}{3}\Big)\times \Big(\frac{-2}{3}\Big)\times \Big(\frac{-2}{3}\Big)\times \Big(\frac{-2}{3}\Big)$
$=\Big(\frac{-2}{3}\Big)^5$

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Question 53 Marks

Simplify the following and express a rational number:
$\Big(\frac{2}{3}\Big)^2\times\Big(\frac{-3}{5}\Big)^3\times \Big(\frac{7}{2}\Big)^2$

Answer

$\Big(\frac{2}{3}\Big)^2\times\Big(\frac{-3}{5}\Big)^3\times \Big(\frac{7}{2}\Big)^2$
$=\frac{2}{3}\times\frac{2}{3}\times\frac{-3}{5}\times\frac{-3}{5}\times\frac{-3}{5}\times\frac{7}{2}\times\frac{7}{2}$
$=\frac{4}{9}\times\frac{-27}{125}\times\frac{49}{4}$
$=\frac{4\times(-27)\times49}{9\times125\times4}$
$=\frac{-3\times49}{125}$
$=\frac{-147}{125}$

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Question 63 Marks

If $2^{\text{n}-7}\times5^{\text{n}-4}=1250,$ find the value of n.

Answer

$2^{\text{n}-7}\times5^{\text{n}-4}=1250$
$\Rightarrow 2^{\text{n}-4-3}\times5^{\text{n}-4}=1250$
$\Rightarrow 2^{\text{n}-4}\times\frac{1}{2^3}\times5^{\text{n}-4}=1250$
$\Rightarrow (2\times5)^{\text{n}-4}=1250\times2^3=1250\times8$
$\Rightarrow 10^{\text{n}-4}=1000=10^4$
Comparing we get,
$\text{n} - 4 = 4$
$\Rightarrow \text{n} = 4 + 4 = 8$
$\therefore \text{n} = 8$

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Question 73 Marks

Simplify:
$\frac{16\times2^{\text{n}+1}-4\times2^{\text{n}}}{16\times2^{\text{n}+2}-2\times2^{\text{n}+2}}$

Answer

$\frac{16\times2^{\text{n}+1}-4\times2^{\text{n}}}{16\times2^{\text{n}+2}-2\times2^{\text{n}+2}}$
$=\frac{2^4\times2^{\text{n}+1}-2^2\times2^{\text{n}}}{2^4\times2^{\text{n}+2}-2\times2^{\text{n}+2}}$
$=\frac{2^{\text{n}+1+4}-2^{\text{n}+2}}{2^{\text{n}+2+4}-2^{\text{n}+2+1}}$
$=\frac{2^{\text{n}+5}-2^{\text{n}+2}}{2^{\text{n}+6}-2^{\text{n}+3}}$
$=\frac{2^\text{n}.2^5-2^\text{n}.2^2}{2^\text{n}.2^6-2^\text{n}.2^3}$
$=\frac{2^\text{n}(2^5-2^2)}{2^\text{n}(2^6-2^3)}$
$=\frac{2^5-2^2}{2^6-2^3}$
$=\frac{32-4}{64-8}=\frac{28}{56}$
$=\frac{1}{2}$

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Question 83 Marks

Express the following in power notation:
$\frac{-1}{128}$

Answer

$\frac{-1}{128}=\frac{(1\times1\times1\times1\times1\times1\times1)}{2\times2\times2\times2\times2\times2\times2 }$
$\begin{array}{c|c} 2 & 128 \\ \hline 2 & 64\\ \hline2&32\\ \hline2&16\\ \hline2&8\\ \hline2&4\\ \hline2&2\\ \hline&1 \end{array}$
$=\frac{(-1)\times(-1)\times(-1)\times(-1)\times(-1)\times(-1)\times(-1)}{2\times2\times2\times2\times2\times2\times2}$
$=\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)\\\times\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)$
$=\Big(\frac{-1}{2}\Big)^7$

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Question 93 Marks

Simplify:
$\frac{16\times2^{\text{n}+1}-8\times2^{\text{n}}}{16\times2^{\text{n}+2}-4\times2^{\text{n}+1}}$

Answer

We have,
$\frac{10\times2^{\text{n}+1}-2^3\times2^\text{n}}{10\times2^{\text{n}+2}-4\times2^{\text{n}+1}}$
$\Rightarrow \frac{2^4\times2^{\text{n}+1}-2^3\times2^\text{n}}{2^4\times2^{\text{n}+2}-2^2\times2^{\text{n}+1}}$
$\Rightarrow \frac{2^3(2^{\text{n}+2}-2^\text{n})}{2^3\times(2^{\text{n}+3}-2^\text{n})}$
$\Rightarrow \frac{2^\text{n}\times2^2-2^\text{n}}{2^\text{n}\times2^3-2^\text{n}}$
$\Rightarrow \frac{2^\text{n}(2^2-1)}{2^\text{n}(2^3-1)}=\frac{4-1}{8-1}$
$=\frac{3}{7}$

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Question 103 Marks

Simplify the following and express a rational number:
$\Big(\frac{-2}{3}\Big)^5\times\Big(\frac{-3}{7}\Big)^3$

Answer

$\Big(\frac{-2}{3}\Big)^5\times\Big(\frac{-3}{7}\Big)^3$
$=\frac{-2}{3}\times\frac{-2}{3}\times\frac{-2}{3}\times\frac{-2}{3}\times\frac{-2}{3}\times\frac{-3}{7}\times\frac{-3}{7}\times\frac{-3}{7}$
$=\frac{-32}{243}\times \frac{-27}{343}=\frac{(-32)\times(-27)}{243\times343}$
$=\frac{32\times1}{9\times343}$
$=\frac{32}{3087}$

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Question 113 Marks

Simplify the following and express a rational number:
$\Bigg\{\Big(\frac{-3}{4}\Big)^3-\Big(\frac{-5}{2}\Big)^3\Bigg\}\times4^2$

Answer

$\Bigg\{\Big(\frac{-3}{4}\Big)^3-\Big(\frac{-5}{2}\Big)^3\Bigg\}\times4^2$
$=\Bigg\{\Big(\frac{-3}{4}\times\frac{-3}{4}\times\frac{-3}{4}\Big)-\Big(\frac{-5}{2}\times\frac{-5}{2}\times\frac{-5}{2}\Big)\Bigg\}\times4\times4$
$=\Big\{\frac{-27}{64}-\frac{-125}{8}\Big\}\times16$
$=\frac{-27+1000}{64}\times16$ (LCM of 64 and 8 = 64)
$=\frac{973}{4}$

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Question 123 Marks

Simplify the following and express a rational number:
$\Big(\frac{-1}{2}\Big)^5\times 2^3\times \Big(\frac{3}{4}\Big)^2$

Answer

$\Big(\frac{-1}{2}\Big)^5\times 2^3\times \Big(\frac{3}{4}\Big)^2$
$=\frac{-1}{2}\times\frac{-1}{2}\times\frac{-1}{2}\times\frac{-1}{2}\times\frac{-1}{2}\\\times2\times2\times2\times\frac{3}{4}\times\frac{3}{4}$
$=\frac{1}{32}\times 8\times \frac{9}{16}$
$=\frac{-1\times9}{4\times16}=\frac{-9}{64}$

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