5 Marks Questions

Question 15 Marks

Using the world population formula $P=6.9\left(1.011^t\right.$, where $t$ is the number of years after 2011 and $P$ and 6.9 billion people in 2011 is the world population in billions of people, estimate :
(a) the population in the year 2050 to the nearest hundred million, and
(b) by what year will the population be double what it was in 2011.

Answer

$\begin{array}{l}\text {(a) Here, } t=2050-2011=39 \\ \therefore \quad P=6.9(1.011)^{39}\end{array}$
Taking $\log$ on both sides, we get
$
\begin{array}{cc}
\Rightarrow& \log _{10} P=\log _{10}\left[6.9(1.011)^{39}\right] \\
\Rightarrow & \log _{10} P=\log _{10}(6.9)+\log _{10}(1.011)^{39} \\
& {\left[\text { Applying rule } \log _a(m n)=\log _a m+\log _a n\right]} \\
\Rightarrow & \log _{10} P=\log _{10}(6.9)+39 \log _{10}(1.011) \\
& {\left[\text { Applying rule } \log _a m^n=n \log _a m\right]} \\
\Rightarrow & \log _{10} P=0.83884+39(0.00475) \\
\Rightarrow & \log _{10} P=0.83884+0.18525 [By\ using\ log\ table] \\
\Rightarrow & \log _{10} P=1.02409 \\
\Rightarrow & \log _{10} P=\text { antilog using log table] }(1.02409) \\
\Rightarrow & P=10.57 \cong 10.6
\end{array}
$
[By using antilog table]
Hence, the population in 2050 will be about 10.6
billion people.

(b) Here,
$P=2(6.9)=13.8$
Now,$13.8=6.9(1.011)^t$
$\Rightarrow$ $2=(1.011)^t$
Taking log on both sides, we get
$
\begin{array}{ll}
\Rightarrow \quad & \log 2=\log (1.011)^t \\
\Rightarrow \quad & \log 2=t \log (1.011)
\end{array}
$$
\begin{array}{ll}
& \text { [Applying rule } \left.\log _a m=n \log _a m\right] \\
\Rightarrow & t=\frac{\log 2}{\log (1.011)} \\
\Rightarrow & t=\frac{0.3010}{0.00475} \\
\Rightarrow & t=63.359 \cong 64
\end{array}
$
Thus, population will be double (i.e., 13.8 billion people) the 2011 population by the year $=2011+$ $64=2075$.

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Question 25 Marks

(i) Solve the following simultaneous logarithmic equations:
$\log _2\left(x y^2\right)=0, \log _2\left(x^2 y\right)=3$
(ii) Solve the following equation for $x$ :
$2 \times 3^{1 / 2 x+2}=23.43$

Answer

Given equations :
$
\begin{array}{l}
\log _2\left(x y^2\right)=0 \ldots(i)\\
\log _2\left(x^2 y\right)=3 \ldots(ii)
\end{array}
$
From eq(i), $\log _2 x+\log _2 y^2=0$
From eq(ii), $\log _2 x^2+\log _2 y=3$
[Applying rule $\log _a(m n)=\log _a m+\log _a n$ ]
or, $\quad \log _2 x+2 \log _2 y=0$
$2 \log _2 x+\log _2 y=3$
[Applying rule $\log _a m^n=n \log _a m$ ]
Let $\log _2 x=X$ and $\log _2 y=Y$
$
\therefore \quad \begin{array}{ll}X+2 Y=0\ldots(iii)\\
2 X+Y=3 \ldots(iv)
\end{array}
$
On solving eqs. (iii) and (iv), we get
$X=2 \text { and } Y=-1$
Therefore, $\log _2 x=2$ and $\log _2 y=-1$
or, $x=2^2 \text { and } y=2^{-1}$
or, $x=4 \text { and } y=\frac{1}{2}$
Hence, values of $x$ and $y$ are 4 and $\frac{1}{2}$, respectively.

(ii) Given,
$
\begin{aligned}
2 \times 3^{\frac{1}{2} x+2} & =23.43 \\
3^{\frac{1}{2} x+2} & =11.715
\end{aligned}
$
Taking $\log$ on both sides, we get
$
\begin{aligned}
& & \log _{10}\left(3^{\frac{1}{2} x+2}\right) & =\log _{10} 11.715 \\
\Rightarrow & & \left(\frac{1}{2} x+2\right) \log _{10} 3 & =\log _{10} 11.715 \\
\Rightarrow & & \frac{1}{2} x+2 & =\frac{\log _{10} 11.715}{\log _{10} 3} \\
\Rightarrow & & \frac{1}{2} x+2 & \cong 2.2399 \quad \quad \text { [Use log table] } \\
\Rightarrow & & \frac{1}{2} x & \cong 0.2399 \\
\Rightarrow & & x & \cong 0.4798 \cong 0.48
\end{aligned}
$

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Question 35 Marks

(i) It is given that $x$ satisfies the logarithm equation $\log _a x=2\left[\log _a k-\log _a 2\right]$,where $k>0, a>0, a \neq 1$.
(a) Find $x$ in terms of $k$, giving the answer in the form not involving logarithm.
Suppose instead that $x$ satisfies
$\log _x(5 y+1)=4+\log _x 3$
where, $x>0, x \neq 1$, and $y>0, y \neq 1$.
(b) Solve the above equation expressing $y$ in terms of $x$, giving the answer in a form not involving logarithm.
(ii) Solve the equation $\frac{1}{6}=\left(\frac{1}{2}\right)^x$ and give your answer as single logarithm of base 2 .

Answer

(i) (a) $
\log _a x=2\left(\log _a k-\log _a 2\right)
$
$
\begin{array}{ll}
\Rightarrow & \log _a x=2 \log _a k-2 \log _a 2 \\
\Rightarrow & \log _a x=\log _a k^2-\log _a 2^2
\end{array}
$
[Applying rule $\log _a m^n=n \log _a m$ ]
$
\Rightarrow \quad \log _a(x)=\log _a \frac{k^2}{4}
$
[Applying rule $\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n$ ]
$
\Rightarrow \quad x=\frac{k^2}{4}
$
[Dropping log from both sides]

$
\begin{array}{l}
\text { (b) }\log _x(5 y+1)=4+\log _x 3 \\
\log _x(5 y+1)=4 \log _x x+\log _x 3 \ {\left[\because \log _a a=1\right]} \\
\Rightarrow \quad \log _x(5 y+1)=\log _x(x)^4+\log _x 3 \\
\text { [Applying rule } \log _a m^n=n \log _a m \text { ] } \\
\Rightarrow \quad \log _x(5 y+1)=\log _x\left(x^4 \times 3\right) \\
\text { [Applying rule } \log _a m^n=n \log _a m \text { ] } \\
\Rightarrow5 y+1=3 x^4 \\
\text { [Dropping log from both sides] } \\
\Rightarrow \quad y=\frac{1}{5}\left(3 x^4-1\right) \text {. }
\end{array}
$

(ii) Given equation,
$
\begin{array}{l}
\Rightarrow \quad\left(\frac{1}{6}\right)=\left(\frac{1}{2}\right)^x \\
\Rightarrow \quad\left(\frac{1}{6}\right)=\left(2^{-1}\right)^x \\
\Rightarrow \quad 6^{-1}=2^{-x} \\
\Rightarrow \quad \log _2 6^{-1}=\log _2 2^{-x} \\
\text { [Taking } \log \text { with base } 2 \text { on both sides] } \\
\Rightarrow \quad-\log _2 6=-x \log _2 2 \\
\text { [Applying rule, } \log _a m^n=n \log _a m \text { ] } \\
\Rightarrow \quad-\log _2 6=-x \quad\left[\because \log _a=1\right] \\
\Rightarrow \quad\quad\quad\quad x=\log _2 6 \\
\Rightarrow \quad\quad\quad\quad x=\log _2(2 \times 3) \\
\Rightarrow \quad\quad\quad\quad x=\log _2 2+\log _2 3 \\
\text { [Applying rule } \left.\log _a(m n)=\log _a m+\log _a n\right] \\
\Rightarrow \quad\quad\quad\quad x=1+\log _2 3\left[\because \log _a a=1\right]
\end{array}
$

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Question 45 Marks

(i) If $\log _a b c=x, \log _b c a=y, \log _c a b=z$, prove that
$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$
(ii) Given that, $p=\log _a 4$ and $q=\log _a 5$Express each of the following logarithms in terms of $p$ and $q$.
(a) $\log _a 100$
(b) $\log _a 0.4$

Answer

(i) $x+1=\log _a b c+\log _{ a } a=\log _a a b c$
$
\begin{array}{l}
y+1=\log _b c a+\log _b b=\log _b a b c \\
z+1=\log _c a b+\log _c c=\log _c a b c
\end{array}
$
Therefore,
$
\begin{aligned}
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1} & =\frac{1}{\log _a a b c}+\frac{1}{\log _b a b c}+\frac{1}{\log _c a b c} \\
& =\log _{a b c} a+\log _{a b c} b+\log _{a b c} c \\
& =\log _{a b c} a b c=1
\end{aligned}
$
Hence proved.

(ii) Given, $p=\log _a 4$ and $q=\log _a 5$
(a)
$
\begin{aligned}
\log _a 100 & =\log _a(25 \times 4) \\
& =\log _a 25+\log _a 4
\end{aligned}
$
[Applying rule $\left.\log _a(m n)=\log _a m+\log _a n\right]$
$\begin{array}{l} \quad\quad\quad\quad =\log _a\left(5^2\right)+\log _a 4 \\ \quad\quad\quad\quad =2 \log _a 5+\log _a 4 \\ \quad\quad\quad\quad{\left[\text { Applying } \log _a m^n=n \log _a m\right]} \\ \quad\quad\quad\quad =2 q+p \quad[\text { Using eq. (i) }] \quad 1\end{array}$

(b)
$
\begin{array}{l}
\begin{aligned}
\log _a(0.4) & =\log _a\left(\frac{2}{5}\right) \\
& =\log _a 2-\log _a 5
\end{aligned} \\
\text { [Applying } \left.\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n\right] \\
\\
\quad\quad\quad\quad=\log _a \sqrt{4}-\log _a 5
\end{array}
$
$
\begin{array}{l}
\quad\quad\quad\quad=\log _a(4)^{1 / 2}-\log _a 5 \\
\quad\quad\quad\quad=\frac{1}{2} \log _a 4-\log _a 5 \\
\quad\quad\quad\quad=\frac{1}{2} p-q .
\end{array}
$
[Applying rule $\left.\log _a m^n=n \log _a m\right] $

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Question 55 Marks

(i) Find the value of
$
\left(\frac{x^b}{x^c}\right)^{(b+c-a)} \cdot\left(\frac{x^c}{x^a}\right)^{(c+a-b)} \cdot\left(\frac{x^a}{x^b}\right)^{(a+b-c)}
$
(ii) If $m$ and $n$ are whole numbers such that $m^n=121$, then find the value of $(m-1)^{n+1}$.
(iii) Simplify : $\frac{(243)^{n / 5} \times 3^{2 n+1}}{9^n \times 3^{n-1}}$.

Answer

(i) Given expression :$
\left(\frac{x^b}{x^c}\right)^{(b+c-a)} \cdot\left(\frac{x^c}{x^a}\right)^{(c+a-b)} \cdot\left(\frac{x^a}{x^b}\right)^{(a+b-c)}
$
$\begin{array}{l}\left.=\left(x^{b-c}\right)^{(b+c-a)} \cdot\left(x^{c-a}\right)^{(c+a-b)} \cdot\left(x^{a-b}\right)\right)^{a+b-c} \\ =x^{(b-c)(b+c-a)} \cdot x^{(c-a)(c+a-b)} \cdot x^{(a-b)(a+b-c)} \\ \quad\left[ Using \left(a^m\right)^n=a^{m n}\right]\end{array}$
$=x^{(b-c)(b+c)-a(b-c)}, x^{(c-a)(c+a)-b(c-a)}, x^{(a-b)(a+b)-c(a-b)}$
$
\begin{array}{l}
=x^{b^2-c^2-a b+a c} \cdot x^{c^2-a^2-b c+a b} \cdot x^{a^2-b^2-c a+b c} \\
=x^{b^2-c^2-a b+a c+c^2-a^2-b c+a b+a^2-b^2-a++b c} \\
{\left[Using\ a^m \cdot a^n=a^{m+n}\right]} \\
=x^0 \\
=1 \text {. } \\{\left[Using\ a^0=1\right] }
\end{array}
$

(ii) We now that, $11^2=121$
$\begin{aligned} & \text { Given, } & m^n & =121\ \ldots(i) \\ & \text { Therefore, } & (11)^2 & =121\ \ldots(ii) \end{aligned}$
On comparing eqs. (i) and (ii), we get
$
m=11 \text { and } n=2
$
Hence, value of $(m-1)^{n+1}=(11-1)^{2+1}$
$
=(10)^3=1000 .
$

(iii) Given expression :
$
\begin{array}{l}
\frac{(243)^{n / 5} \times 3^{2 n+1}}{9^n \times 3^{n-1}} \\
=\frac{\left(3^5\right)^{n / 5} \times 3^{2 n+1}}{\left(3^2\right)^n \times 3^{n-1}} \\
=\frac{3^n \times 3^{2 n+1}}{3^{2 n} \times 3^{n-1}} \quad\left[\text { Using }\left(a^m\right)^n=a^{m n n}\right] \\
=\frac{3^{n+2 n+1}}{3^{2 n+n-1}} \quad\left[\text { Using } a^m \cdot a^n=a^{m+n}\right] \\
=(3)^{(3 n+1)-(3 n-1)} \quad \\
=(3)^2=9
\end{array}
$

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Question 65 Marks

(i) Simplify $: \sqrt[6]{a^{4 b} x^6} \cdot\left(a^{\frac{2}{3}} x^{-1}\right)^{-b}$.
(ii) Find the value of $k$ from $(\sqrt{9})^{-7} \times(\sqrt{3})^{-5}=3^k$.
(iii) Mahima says for any number $a, a^2$ is less than $a^3$. For example, when $a=3,3^2<3^3$ as $9<27$. Find an example to show that Mahima is wrong.

Answer

$
\begin{array}{l}
\text { (i) } \sqrt[6]{a^{4 b} \cdot x^6} \cdot\left(a^{\frac{2}{3}} \cdot x^{-1}\right)^{-b} \\
=\left(a^{4 b} x^6\right)^{\frac{1}{6}} \cdot\left(a^{\frac{2}{3}}\right)^{-b} \cdot\left(x^{-1}\right)^{-b} \quad\left[\text { Using } \sqrt[4]{a}=(a)^{\frac{1}{r}}\right]
\end{array}
$
$\begin{array}{l}=\left(a^{4 b}\right)^{\frac{1}{6}} \cdot\left(x^6\right)^{\frac{1}{6}} \cdot a^{\frac{-2 b}{3}} \cdot x^b \\ =a^{\frac{2}{3} b} \cdot x \cdot a^{\frac{-2}{3} b} \cdot x^b \quad \quad\left[\text { Using }\left(a^m\right)^n=a^{m n}\right]\end{array}$
$\begin{array}{ll}=a^{\frac{2}{3} b-\frac{2}{3} b} \cdot x^{1+b} & {\left[\text { Using } a^m \cdot a^n=a^{m+n}\right]} \\ =a^0 \cdot x^{1+b} & \end{array}$
$\begin{array}{lr}=1 \cdot x^{1+b} & {\left[\text { Using } a^0=1\right]} \\ =x^{1+b} . \end{array}$

$\begin{array}{l}\text { (ii) Given, }(\sqrt{9})^{-7} \times(\sqrt{3})^{-5}=3^k \\ \text { or, }\left(3^{2 \times \frac{1}{2}}\right)^{-7} \times\left(3^{\frac{1}{2}}\right)^{-5}=3^k \quad\left[\text { Using } \sqrt[r]{a}=(a)^{\frac{1}{r}}\right]\end{array}$
or, $3^{-7-\frac{5}{2}}=3^k$
or, $3^{\frac{19}{2}}=3^k$
Since, equal base has equal power.
Therefore,$k=\frac{-19}{2}$

(iii) Mahima is right for all numbers bigger than 1, but she should be wrong for any numbers include negative numbers.
For example :
$
\begin{aligned}
a =-3 \\
(-3)^2 =9
\end{aligned}
$
$(-3)^3=-27$
Therefore, $\quad 9>-27 \Rightarrow a^2>a^3$.
Another example : $a=\frac{1}{2}$
$\left(\frac{1}{2}\right)^2=\frac{1}{4}$
and $\left(\frac{1}{2}\right)^3=\frac{1}{8}$
Therefore, $\frac{1}{4}>\frac{1}{8} \Rightarrow a^2>a^3$.

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Question 75 Marks

(i) Evaluate $5^{-2} \times 100^{0.5}$. Give your answer in its simplest form.
(ii) Express $5^{20}$ as the power of 25 .
(iii) Show that: $27^{\frac{2}{3}}=\frac{1}{9}$.
(iv) Evaluate : $\left(\frac{125}{27}\right)^{\frac{-1}{3}}$.
(v) Evaluate (81) ${ }^{0.5} \times 6^{-2}$. Give your answer in simplest form.
(vi) Simplify : $\left(\frac{27}{8}\right)^{\frac{-2}{3}}$.

Answer

(i) $(5)^{-2} \times(100)^{0.5}$
Here, $5^{-2}=\frac{1}{5^2}=\frac{1}{25}\left[\text { Using } a^{-n}=\frac{1}{a^n}\right]$
and $(100)^{0.5}=(100)^{\frac{1}{2}}=\sqrt{100}=10$
Therefore,
$(5)^{-2} \times(100)^{0.5}=\frac{1}{25} \times 10=\frac{10}{25}=\frac{2}{5}$.

(ii) $5^{20}$
$
\begin{aligned}
5^{20} & =(\sqrt{25})^{20}=\left[(25)^{\frac{1}{2}}\right]^{20} \\
& =(25)^{20 \times \frac{1}{2}} \quad\left[\because\left(a^{m 1}\right)^n=a^{m n n}\right] \\
& =(25)^{10}
\end{aligned}
$
$
\text { Therefore, } \quad 5^{20}=(25)^{10} \text {. }
$

(iii) To show $27^{-\frac{2}{3}}=\frac{1}{9}$
We split $\frac{-2}{3}$ in $\frac{1}{3} \times(-2)$.
$\begin{aligned} \text { L.H.S. }=(27)^{\frac{2}{3}} & =\left[(27)^{\frac{1}{3}}\right]^{-2} \\ & =[\sqrt[3]{27}]^{-2}\end{aligned}$
$\begin{array}{l}\quad\left[\text { Using } \sqrt[r]{a}=(a)^{\frac{1}{r}}\right] \\ =3^{-2} \\ =\frac{1}{3^2}=\frac{1}{9}=\text { R.H.S. }\end{array}$

(iv) $\left(\frac{125}{27}\right)^{\frac{-1}{3}}$
$\begin{aligned}\left(\frac{125}{27}\right)^{\frac{-1}{3}} & =\left(\frac{27}{125}\right)^{\frac{1}{3}} \\ & =\sqrt[3]{\frac{27}{125}} \\ & =\frac{\sqrt[3]{27}}{\sqrt[3]{125}}=\frac{3}{5}\end{aligned}$

(v) $(81)^{0.5} \times 6^{-2}$
Here, $(81)^{0.5}=(81)^{\frac{1}{2}}=\left(9^2\right)^{\frac{1}{2}}=9$
and $6^{-2}=\frac{1}{6^2}=\frac{1}{36}$
Therefore, $(81)^{0.5} \times 6^{-2}=9 \times \frac{1}{36}=\frac{1}{4}$.

(vi) $\left(\frac{27}{8}\right)^{\frac{-2}{3}}$
$\begin{aligned}\left(\frac{27}{8}\right)^{\frac{-2}{3}}=\left(\frac{8}{27}\right)^{\frac{2}{3}} & =\left[\left(\frac{8}{27}\right)^{\frac{1}{3}}\right]^2 =\left[\left(\frac{2^3}{3^3}\right)^{\frac{1}{3}}\right]^2\end{aligned}$
$=\left[\left\{\left(\frac{2}{3}\right)^3\right\}^{\frac{1}{3}}\right]^2 \quad\left[\because\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}\right]$
$=\left[\left(\frac{2}{3}\right)^{3 \times \frac{1}{3}}\right]^2$
$\left[\right.$ Using $\left.\left(a^{m m}\right)^n=a^{m n}\right]$
$=\left(\frac{2}{3}\right)^2=\frac{4}{9}$.

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Question 85 Marks

(i) Find the value of $\left(\frac{x^a}{x^b}\right)^{a+b} \times\left(\frac{x^b}{x^c}\right)^{b+c} \times\left(\frac{x^c}{x^a}\right)^{c+a}$
(ii) If $a^x=b, b^y=c, c^z=a$, then find value of $x y z$.
(iii) Find the value of
$\left[\left\{(2)^{1 / 2} \cdot(4)^{3 / 4} \cdot(8)^{5 / 6} \cdot(16)^{7 / 8} \cdot(32)^{9 /10}\right\}^4\right]^{4 / 25}$

Answer

(i) Given, $\left(\frac{x^a}{x^b}\right)^{a+b} \times\left(\frac{x^b}{x^c}\right)^{b+c} \times\left(\frac{x^c}{x^a}\right)^{c+a}$
$\begin{array}{l}=\left(x^{a-b}\right)^{a+b} \times\left(x^{b-c}\right)^{b+c} \times\left(x^{c-a}\right)^{c+a} \\ =x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}\end{array}$
$\begin{array}{l}\quad\left[\because m^2-n^2=(m+n)(m-n)\right] \\ =x^{a^2-b^2+b^2-c^2+c^2-a^2} \\ \quad\left[ U \operatorname{sing} a^{m \prime} a^n=a^{m+n}\right]\end{array}$
$=x^0=1$.

(ii) Given, $\quad a^x=b, b^y=c$ and $c^z=a$
Since, $\quad a^x=b$
$\begin{aligned} \Rightarrow & \quad\left(a^x\right)^y=b^y & \\ \Rightarrow &\quad a^{x y}=b^y & {\left[\text { Using }\left(a^m\right)^n=a^{m n}\right] }\end{aligned}$
$\Rightarrow \quad a^{n y}=c$ [Given $b^y=c$ ]
$\Rightarrow \quad\left(a^{x y}\right)^z=c^z$
$\begin{array}{lll}\Rightarrow & a^{x y z}=a & {\left[\text { Given } c^z=a\right]} \\ \Rightarrow & a^{x y z}=a^1 & \end{array}$
Since, same base as equal power, therfore
$x y z=1$.

(iii) $\left[\left\{(2)^{\frac{1}{2}} \cdot(4)^{\frac{3}{4}} \cdot(8)^{\frac{5}{6}} \cdot(16)^{\frac{7}{8}} \cdot(32)^{\frac{9}{10}}\right\}^4\right]^{\frac{4}{25}}$
$=\left[\left\{(2)^{\frac{1}{2}} \cdot\left(2^2\right)^{\frac{3}{4}} \cdot\left(2^3\right)^{\frac{5}{6}} \cdot\left(2^4\right)^{\frac{7}{8}} \cdot\left(2^5\right)^{\frac{9}{10}}\right\}^4\right]^{\frac{4}{25}}$
$\begin{array}{r}=\left[\left\{(2)^{\frac{1}{2}} \cdot(2)^{\frac{3}{2}} \cdot(2)^{\frac{5}{2}} \cdot(2)^{\frac{7}{2}} \cdot(2)^{\frac{9}{2}}\right\}^4\right]^{\frac{4}{25}} \\ {\left[\text { Using }\left(a^{m}\right)^n=a^{m n}\right]}\end{array}$
$=\left[\left(2^{\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+\frac{7}{2}+\frac{9}{2}}\right)^4\right]^{\frac{4}{25}}$[Using $\left.a^m \cdot a^n=a^{m+n}\right]$
$=\left\{\left(2^{\frac{25}{2}}\right)^4\right\}^{\frac{4}{25}}$
$\begin{array}{l}=\left(2^{25 \times 2}\right)^{\frac{4}{25}} \\ =2^{4 \times 2}=2^8=256\end{array}$

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Applied Maths 5 Marks Questions

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