4 Mark Question

Question 14 Marks

Prove that:$\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}=\frac{16}{3}$

Answer

We have to prove that $\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}=\frac{16}{3}$
Now,
$\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}\\=\dfrac{1^{-2}}{4^{-2}}-3\times2^{3\times\frac{2}{3}}\times4^0+\frac{3^{2\times-\frac{1}{2}}}{4^{2\times-\frac{1}{2}}}$
$\Rightarrow\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}\\=\frac{1}{2^{-4}}-3\times2^2\times4^0+\frac{3^{-1}}{2^{-2}}$
$\Rightarrow\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}\\=\frac{1}{\frac{1}{2^4}}-3\times2^2\times4^0+\frac{\frac{1}{3}}{\frac{1}{2^2}}$
$\Rightarrow\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}\\=1\times\frac{2^4}{1}-3\times2^2\times1+\frac{1}{3}\times\frac{2^2}{1}$
$\Rightarrow\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}\\=\frac{16}{1}-\frac{12}{1}+\frac{4}{3}=\frac{16}{3}$
Hence, $\Big(\frac{1}{4}\Big)^{-2}-3\times8^{\frac{2}{3}}\times4^0+\Big(\frac{9}{16}\Big)^{-\frac{1}{2}}=\frac{16}{3}$

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Question 24 Marks

Solve the following equations:$8^{\text{x}+1}=16^{\text{y}+2}$ and $\Big(\frac{1}{2}\Big)^{3+\text{x}}=\Big(\frac{1}{4}\Big)^{3\text{y}}$

Answer

$8^{\text{x}+1}=16^{\text{y}+2}$$\Rightarrow(2^3)^{\text{x}+1}=(2^4)^{\text{y}+2}$
$\Rightarrow2^{3\text{x}+3}=2^{\text{4y}+8}$
$\Rightarrow3\text{x}-4\text{y}=5\ ...(1)$
And $\Big(\frac{1}{2}\Big)^{3+\text{x}}=\Big(\frac{1}{4}\Big)^{3\text{y}}$
$\Rightarrow(2^{-1})^{3+\text{x}}=\Big(\frac{1}{2^2}\Big)^{3\text{y}}$
$\Rightarrow2^{-3-\text{x}}=(2^{-2})^{3\text{y}}$
$\Rightarrow-3-\text{x}=-6\text{y}$
$\Rightarrow\text{x}-6\text{y}=-3\ ...(2)$
Multiplying eqn (2) by 3, we get
$3\text{x}-18\text{y}=-9\ ...(3)$
Subtracting eqn (3) from (1), we get
$14\text{y}=14\Rightarrow\text{y}=1$
$\Rightarrow3\text{x}-4(1)=5\ ...[\text{From (1)}]$
$\Rightarrow3\text{x}=9\Rightarrow\text{x}=3$
Hence, $\text{x}=3$ and $\text{y}=1$

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Question 34 Marks

For any positive real number x, find the value of $\Big(\frac{\text{x}^{\text{a}}}{\text{x}^{\text{b}}}\Big)^{\text{a}+\text{b}}\times\Big(\frac{\text{x}^{\text{b}}}{\text{x}^{\text{c}}}\Big)^{\text{b}+\text{c}}\times\Big(\frac{\text{x}^{\text{c}}}{\text{x}^{\text{a}}}\Big)^{\text{c}+\text{a}}$

Answer

We have to find the value of $\text{L}=\Big(\frac{\text{x}^{\text{a}}}{\text{x}^{\text{b}}}\Big)^{\text{a}+\text{b}}\times\Big(\frac{\text{x}^{\text{b}}}{\text{x}^{\text{c}}}\Big)^{\text{b}+\text{c}}\times\Big(\frac{\text{x}^{\text{c}}}{\text{x}^{\text{a}}}\Big)^{\text{c}+\text{a}}$$\text{L}=\Big(\frac{\text{x}^{\text{a}(\text{a}+\text{b})}}{\text{x}^{\text{b}(\text{a}+\text{b})}}\Big)\times\Big(\frac{\text{x}^{\text{b}(\text{b}+\text{c})}}{\text{x}^{\text{c}(\text{b}+\text{c})}}\Big)\times\Big(\frac{\text{x}^{\text{c}(\text{c}+\text{a})}}{\text{x}^{\text{a}(\text{c}+\text{a})}}\Big)$
$=\frac{\text{x}^{\text{a}^{2}+\text{ab}}}{\text{x}^{\text{ba}+\text{b}^2}}\times\frac{\text{x}^{\text{b}^{2}+\text{bc}}}{\text{x}^{\text{bc}+\text{c}^2}}\times\frac{\text{x}^{\text{c}^{2}+\text{ca}}}{\text{x}^{\text{ac}+\text{a}^2}}$
By using rational exponents $\text{a}^\text{m}\times\text{a}^\text{n}=\text{a}^{\text{m}+\text{n}}$ we get$\text{L}=\frac{\text{x}^{\text{a}^2+\text{ab+b}^2+\text{bc+c}^2+\text{ca}}}{\text{x}^{\text{ab}+\text{b}^2+\text{bc}+\text{c}^2+\text{ac+a}^2}}$
By using rational exponents $\frac{\text{a}^\text{m}}{\text{a}^{\text{n}}}=\text{a}^{\text{m+n}}$ we get$\text{L}=\text{x}^{(\text{a}^2+\text{ab}+\text{b}^2+\text{bc}+\text{c}^2+\text{ca})-(\text{ab}+\text{b}^2+\text{bc}+\text{c}^2+\text{ac}+\text{a}^2)}$
$\text{L}=\text{x}^{\text{a}^2+\text{ab}+\text{b}^2+\text{bc}+\text{c}^2+\text{ca}-\text{ab}+\text{b}^2+\text{bc}+\text{c}^2+\text{ac}+\text{a}^2}$
$=\text{x}^0$
By definition we can write x° as 1 Hence the value of expression is 1.

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Question 44 Marks

Simplify $\Big[\Big\{(625)^{\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Big]^2$

Answer

We have to simplify$\bigg[\Big\{(625)^{\frac{1}{2}}\Big\}^{-\frac{1}{4}}\bigg]^2$. So,$\Bigg[\Big\{(625)^{\frac{1}{2}}\Big\}^{-\frac{1}{4}}\bigg]^2=\Bigg[\Big\{\frac{1}{625^{\frac{1}{2}}}\Big\}^{\frac{-1}{4}}\Bigg]^2$
$=\Bigg[\Big\{\frac{1}{5^{4\times\frac{1}{2}}}\Big\}^{\frac{-1}{4}}\Bigg]^2$
$=\Bigg[\Big\{\frac{1}{5^2}\Big\}^{\frac{-1}{4}}\Bigg]^2$
$=\bigg[\Big\{\frac{1}{5^{2\times\frac{-1}{4}}}\Big\}\bigg]^2$
$\Bigg[\Big\{(625)^{\frac{-1}{2}}\Big\}^{\frac{-1}{4}}\Bigg]^2=\Bigg[\bigg\{\frac{1}{5^{\frac{-1}{2}}}\bigg\}\Bigg]^2$
$\bigg[\begin{Bmatrix}\frac{1}{\frac{1}{5^{\frac{1}{2}}}} \end{Bmatrix}\bigg]^2$
$=\bigg[\Big\{1\times5^{\frac{1}{2}}\Big\}\bigg]^2$
$=5^{\frac{1}{2}\times2}=5$
Hence, the value of $\Bigg[\Big\{(625)^{\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2$ is 5.

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Question 54 Marks

Simplify:$\sqrt[\text{lm}]{\frac{\text{x}^\text{l}}{\text{x}^\text{m}}}\times\sqrt[\text{mn}]{\frac{\text{x}^\text{m}}{\text{x}^\text{n}}}\times\sqrt[\text{nl}]{\frac{\text{x}^\text{n}}{\text{x}^\text{l}}}$

Answer

$\sqrt[\text{lm}]{\frac{\text{x}^\text{l}}{\text{x}^\text{m}}}\times\sqrt[\text{mn}]{\frac{\text{x}^\text{m}}{\text{x}^\text{n}}}\times\sqrt[\text{nl}]{\frac{\text{x}^\text{n}}{\text{x}^\text{l}}}$$=\Big(\frac{\text{x}^{\text{l}}}{\text{x}^\text{m}}\Big)^{\frac{1}{\text{lm}}}\times\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\frac{1}{\text{mn}}}\times\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\frac{1}{\text{nl}}}$
$=\big(\text{x}^{\text{l}-\text{m}}\big)^{\frac{1}{\text{lm}}}\times\big(\text{x}^{\text{m}-\text{n}}\big)^{\frac{1}{\text{mn}}}\times\big(\text{x}^{\text{n}-\text{}1}\big)^{\frac{1}{\text{nl}}}$
$=\text{x}^{\frac{\text{l}-\text{m}}{\text{mn}}}\times\text{x}^{\frac{\text{m}-\text{n}}{\text{mn}}}\times\text{x}^{\frac{\text{n}-\text{l}}{\text{nl}}}$
$=\text{x}^{\frac{\text{l}-\text{m}}{\text{mn}}+\frac{\text{m}-\text{n}}{\text{mn}}+\frac{\text{n}-\text{l}}{\text{nl}}}$
$=\text{x}^{\frac{\text{n}(\text{l}-\text{m})+\text{l}(\text{m}-\text{n})+\text{m}(\text{n}-\text{l})}{\text{lmn}}}$
$=\text{x}^{\frac{\text{n}\text{l}-\text{m}+\text{l}\text{m}-\text{n}\text{l}+\text{m}\text{n}-\text{lm}}{\text{lmn}}}$
$=\text{x}^0$
$=\text{1}$

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Question 64 Marks

For any positive real number x, write the value of $\big\{(\text{x}^\text{a})^\text{b}\big\}^{\frac{1}{\text{ab}}}\big\{(\text{x}^\text{b})^\text{c}\big\}^{\frac{1}{\text{bc}}}\big\{(\text{x}^\text{c})\text{a}\big\}^{\frac{1}{\text{ca}}}$

Answer

We have to simplify $\big\{(\text{x}^\text{a})^\text{b}\big\}^{\frac{1}{\text{ab}}}\big\{(\text{x}^\text{b})^\text{c}\big\}^{\frac{1}{\text{bc}}}\big\{(\text{x}^\text{c})\text{a}\big\}^{\frac{1}{\text{ca}}}$ So, $\big\{(\text{x}^\text{a})^\text{b}\big\}^{\frac{1}{\text{ab}}}\big\{(\text{x}^\text{b})^\text{c}\big\}^{\frac{1}{\text{bc}}}\big\{(\text{x}^\text{c})\text{a}\big\}^{\frac{1}{\text{ca}}}\\=\big\{\text{x}^{\text{ab}}\big\}^{\frac{1}{\text{ab}}}\big\{\text{x}^{\text{bc}}\big\}^{\frac{1}{\text{bc}}}\big\{\text{x}^{\text{ca}}\big\}^{\frac{1}{\text{ca}}}$ $=\text{x}^{\text{ab}\times\frac{1}{\text{ab}}}\times\text{x}^{\text{bc}\times\frac{1}{\text{bc}}}\times\text{x}^{\text{ca}\times\frac{1}{\text{ca}}}$ $=\text{x}^1\times\text{x}^1\times\text{x}^1$ By using rational exponents $\text{a}^\text{m}\times\text{a}^\text{n}=\text{a}^{\text{m}+\text{n}},$ we get$\Big\{\big(\text{x}^{\text{a}}\big)^\text{b}\Big\}^{\frac{1}{\text{ab}}}\Big\{\big(\text{x}^\text{b}\big)^\text{c}\Big\}^{\frac{1}{\text{bc}}}\Big\{\big(\text{x}^\text{c}\big)^\text{a}\Big\}^{\frac{1}{\text{ca}}}=\text{x}^{1+1+1}$
$=\text{x}^3$
Hence the value of $\big\{(\text{x}^\text{a})^\text{b}\big\}^{\frac{1}{\text{ab}}}\big\{(\text{x}^\text{b})^\text{c}\big\}^{\frac{1}{\text{bc}}}\big\{(\text{x}^\text{c})\text{a}\big\}^{\frac{1}{\text{ca}}}$ is $\text{x}^3$

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Question 74 Marks

If $(x-1)^3=8$, What is the value of $(x+1)^2$ ?

Answer

We have to find the value of $(x+1)^2$, where $(x-1)^3=8$
Consider $(x-1)^3=2^3$
By equating the base, we get
$x-1=2$
$x=2+1$
$x=3$
By substituting $x=3$ in $(x+1)^2$
$=(x+1)^2$
$=(3+1)^2$
$=4^2$
$=4 \times 4$
$=16$
Hence the value of $(x+1)^2$ is 16 .

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Question 84 Marks

Simplify:$\Big(\frac{\text{x}^{\text{a}+\text{b}}}{\text{x}^\text{c}}\Big)^{\text{a}-\text{b}}\Big(\frac{\text{x}^{\text{b}+\text{c}}}{\text{x}^\text{a}}\Big)^{\text{b}-\text{c}}\Big(\frac{\text{x}^{\text{c}+\text{a}}}{\text{x}^\text{b}}\Big)^{\text{c}-\text{a}}$

Answer

$\Big(\frac{\text{x}^{\text{a}+\text{b}}}{\text{x}^\text{c}}\Big)^{\text{a}-\text{b}}\Big(\frac{\text{x}^{\text{b}+\text{c}}}{\text{x}^\text{a}}\Big)^{\text{b}-\text{c}}\Big(\frac{\text{x}^{\text{c}+\text{a}}}{\text{x}^\text{b}}\Big)^{\text{c}-\text{a}}$$=\big(\text{x}^{\text{a}+\text{b}+\text{c}}\big)^{\text{a}-\text{b}}\big(\text{x}^{\text{b}+\text{c}+\text{a}}\big)^{\text{b}-\text{c}}\big(\text{x}^{\text{c}+\text{a}-\text{b}}\big)^{\text{c}-\text{a}}$
$=\text{x}^{(\text{a}-\text{b})(\text{a}+\text{b}-\text{c})}\times\text{x}^{(\text{b}-\text{c})(\text{b}+\text{c}-\text{a})}\times\text{x}^{(\text{c}-\text{a})(\text{c}+\text{a}-\text{b})}$
$=\text{x}^{\text{a}^\text{2}+\text{ab}-\text{ac}-\text{ab}-\text{b}^\text{2}+\text{bc}}\times\text{x}^{\text{b}^\text{2}+\text{bc}-\text{ab}-\text{bc}-\text{c}^2+\text{ac}}\times\text{x}^{\text{c}^2+\text{ac}-\text{bc}-\text{ac}-\text{a}^\text{2}+\text{ab}}$
$=\text{x}^{\text{a}^2-\text{ac}-\text{b}^2+\text{bc}}\times\text{x}^{\text{b}^2-\text{ab}-\text{c}^2+\text{ac}}\times\text{x}^{\text{c}^2-\text{bc}-\text{a}^2+\text{ab}}$
$=\text{x}^{\text{a}^2-\text{ac}-\text{b}^2+\text{bc}+\text{b}^2-\text{ab}-\text{c}^2+\text{ac}+\text{c}^2-\text{bc}-\text{a}^2+\text{ab}}$
$=\text{x}^0$
$=1$

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