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Applied Maths 2 Marks Questions

Indices and Logarithms · CBSE STD 11 Science (CBSE - English Medium). 17 questions.

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Question 12 Marks
Find the value of $\log 5$, if given $\log 2=0.3010$.
Answer

$
\begin{array}{l}
\qquad \begin{array}{l} 
\log 5=\log \frac{10}{2}=\log 10-\log 2 \\
\text { [Applying rule } \left.\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n\right] \\
=1-0.3010 \quad[\because \log 10=1 \\
\text { and given } \log 2=0.3010] \\
=0.6990
\end{array}
\end{array}
$
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Question 22 Marks
Find the logarithm of 64 to the base $2 \sqrt{2}$.
Answer

$
\begin{aligned}
\log _{2 \sqrt{2}} 64= & \log _{2 \sqrt{2}} 8^2 \\
= & 2 \log _{2 \sqrt{2}} 8 \quad[\text { Applying rule } \left.\ \log _a m^n=n \log _a^m\right] \\
= & 2 \log _{2 \sqrt{2}}(2 \sqrt{2})^2
\end{aligned}
$
$
\begin{array}{l}
\quad\quad\quad\quad= 4 \log _{2 \sqrt{2}}(2 \sqrt{2}) \\
\quad\quad\quad\quad=4 \times 1=4
\end{array}
$
[Applying rule $\log _a a=1$ ]
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Question 32 Marks
Find the value of $\log _2 \log _2 \log _2 16$.
Answer
We have,
$\log _2 \log _2 \log _2 16$
$=\log _2 \log _2\left(\log _2 2^4\right)$
$=\log _2 \log _2\left(4 \log _2 2\right)$
[Applying rule $\log _a m^n=n \log _a{ }^m$ ]
$\begin{array}{l}=\log _2 \log _2 4\left[\text { Applying rule } \log _a a=1\right] \\ =\log _2 \log _2\left(2^2\right) \\ =\log _2\left(2 \log _2 2\right) \\ =\log _2 2 \\ =1\end{array}$
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Question 42 Marks
If $\log x+\log y=\log (x+y)$, then find the relation between $x$ and $y$.
Answer
Given,
$
\begin{array}{l}
\log x+\log y=\log (x+y) \\
\Rightarrow \quad \log (x, y)=\log (x+y) \\
\text { [Applying rule } \left.\log _a(m n)=\log _a m+\log _a n\right]
\end{array}
$
$\begin{aligned} \Rightarrow x y =x+y \\ \Rightarrow x y-y =x \\ \Rightarrow y(x-1) =x \\ \Rightarrow ; y =\frac{x}{x-1}\end{aligned}$
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Question 52 Marks
Simplify : $\left(x^a y^{-b}\right)^3\left(x^3 y^2\right)^{-a}$.
Answer
$\left(x^a y^{-b}\right)^3\left(x^3 y^2\right)^{-a}$
$\begin{array}{lr}=\left(x^a\right)^3 \cdot\left(y^{-b}\right)^3 \cdot\left(x^3\right)^{-a} \cdot\left(y^2\right)^{-a} & \\ =x^{3 a} \cdot y^{-3 b} \cdot x^{-3 a} \cdot y^{-2 a} & {\left[Using\left(a^m\right)^n=a^{m n}\right]} \\ =x^{3 a-3 a} \cdot y^{-2 a-3 b} & \\ =x^0 \cdot y^{-(2 a+3 b)} & {\left[ \text { Using } a^m \cdot a^n=a^{m+n}\right]} \\ =1 \cdot y^{-(2 a+3 b)} & {\left[ Using\ a^0=1\right]} \\ =\frac{1}{y^{2 a+3 b}} . & {\left[ Using\ a^{-r}=\frac{1}{a^r}\right]}\end{array}$
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Question 62 Marks
Simplify : $y^{a-b} \times y^{b-c} \times y^{c-a} \times y^{-a-b}$
Answer
$y^{a-b} \times y^{b-c} \times y^{c-a} \times y^{-a-b}$
$=y^{a-b+b-c+c-a-a-b}$
$\left[\right.$ Using $a^m \times a^n=a^{m+n}$ ]
$=y^{-a-b}$
$=\frac{1}{y^{a+b}}$ $\left[U \operatorname{sing} x^{-n}=\frac{1}{x^n}\right]$
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Question 72 Marks
Find value of
$
\left\{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2\right\} \div\left\{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3\right\}
$
Answer
$\begin{aligned} \frac{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2}{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3} & =\frac{3^6 \times 4^6 \times 5^6}{3^6 \times 4^6 \times 5^6} \\ & =1 \quad\left[\text { Using }\left(a^{m}\right)^n=a^{m n}\right]\end{aligned}$
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Question 82 Marks
If $\log (0.57)=\overline{1} .756$, then find the value of $\log 57+\log (0.57)^3+\log \sqrt{0.57}$
Answer

$\begin{array}{rlrl}\text { log }(0.57) & =\overline{1} .756 \\ \Rightarrow \quad \log 57 & =1.756 \\ & & {[\because \text { mantissa will remain same }]}\end{array}$
$
\begin{array}{l}
\therefore \log 57+\log (0.57)^3+\log \sqrt{0.57} \\
=\log 57+3 \log \left(\frac{57}{100}\right)+\log \left(\frac{57}{100}\right)^{1 / 2}
\end{array}
$
[Applying rule $\log _a m^n=n \log _a m$ ]
$\begin{array}{r}=\log 57+3 \log 57-3 \log 100+\frac{1}{2} \log 57-\frac{1}{2} \log 100 \\ {\left[\text { Applying rule } \log _a \frac{m}{n}=\log _a m-\log _a n\right]}\end{array}$
$\begin{array}{l}=\frac{9}{2} \log 57-\frac{7}{2} \log 100 \\ =\frac{9}{2} \times 1.756-\frac{7}{2} \times 2 \quad\left[\because \log _{10} 100=2\right]\end{array}$
$\begin{array}{l}=7.902-7 \\ =0.902 .\end{array}$
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Question 92 Marks
$\log _{10} 2=x$ and $\log _{10} 3=y$, then find the value of $\log _{10} 60$.
Answer
Given, $\log _{10} 2=x$ and $\log _{10} 3=y$ 
We know that,
$\log _a(m n)=\log _a m+\log _a n$
Therefore,
$
\begin{aligned}
\log _{10}(60) & =\log _{10}(6 \times 10) \\
& =\log _{10} 6+\log _{10} 10 \\
& =\log (2 \times 3)+1 \\
& \quad\left[\because \log _{10} 10=1\right] \\
& =\log _{10} 2+\log _{10} 3+1 \\
& =x+y+1
\end{aligned}
$
Hence, value of $\log _{10} 60=x+y+1$.
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Question 102 Marks
If $2 \log x=4 \log 3$, then find the value of $x$.
Answer
Given,
$2 \log x=4 \log 3$
$\Rightarrow \quad \log x^2=\log 3^4 \quad \begin{array}{r}\text { [Applying rule }\left.\log _a m^n=n \log _a m\right]\end{array}$
$\begin{array}{ll}\Rightarrow & x^2=3^4 \\ \Rightarrow & x=\sqrt{3^4} \\ \Rightarrow & x=\left(3^4\right)^{1 / 2} \\ \Rightarrow & x=3^2=9\end{array}$
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Question 112 Marks
Find the number whose logarithm is 2.4678 .
Answer
From antilog table, for mantissa 0.467 , the number $=2931$
for mean difference 8 , the number $=5$
For mantissa 0.4678 , the number $=2931+5=2936$
The characteristic is 2 , therefore the number must have 3 digits in the integral part
Hence, Antilog $2.4678=293.6$
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Question 122 Marks
Solve : $\frac{1}{2} \log _{10} 25-2 \log _{10} 3+\log _{10} 18$
Answer
The given expression is
$
\begin{array}{l} 
\frac{1}{2} \log _{10} 25-2 \log _{10} 3+\log _{10} 18 \\
= \log _{10}(25)^{1 / 2}-\log _{10}(3)^2+\log _{10} 18 \\
{\left[\text { Applying rule } \log _9(m)^n=n \log _d m\right] }
\end{array}
$
$
\begin{array}{l}
=\log _{10} 5-\log _{10} 9+\log _{10} 18 \\
=\log _{10}\left(\frac{5}{9}\right)+\log _{10} 18
\end{array}
$
[Applying rule $\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a{ }^n$ ]
$\begin{array}{l}=\log _{10}\left(\frac{5 \times 18}{9}\right) \\ \text { }\left[\text { Applying rule } \log _a(m n)=\log _a m+\log _a n\right] \\ \left.=\log _{10} 10 \ =1 \quad \quad [\text { Applying rule } \log _a a=1\right] \\ \end{array}$
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Question 132 Marks
Simplify : $\frac{2 a^{\frac{1}{2}} \times a^{\frac{2}{3}} \times 6 a^{\frac{-7}{3}}}{9 a^{\frac{-5}{3}} \times a^{\frac{3}{2}}}$, if $a=4$.
Answer
$\frac{2 a^{\frac{1}{2}} \cdot a^{\frac{2}{3}} \times 6 a^{\frac{-7}{3}}}{9 a^{\frac{-5}{3}} \times a^{\frac{3}{2}}}=\frac{2.2 .3 \cdot a^{\frac{1}{2}+\frac{2}{3}-\frac{7}{3}}}{3.3 a^{\frac{-5}{3}+\frac{3}{2}}}$
$=\frac{4}{3} \frac{a^{\frac{3+4-14}{6}}}{a^{\frac{-10+9}{6}}}$
$=\frac{4}{3} \frac{a^{\frac{-7}{6}}}{a^{\frac{-1}{6}}}$
$=\frac{4}{3} a^{\frac{-7}{6}+\frac{1}{6}}$
$\begin{array}{l}=\frac{4}{3} a^{\frac{-6}{6}} \\ =\frac{4}{3} a^{-1}\end{array}$
$=\frac{4}{3} \cdot \frac{1}{a}$
$=\frac{4}{3} \cdot \frac{1}{4}$[Given $a=4$ ]
$=\frac{1}{3}$.
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Question 152 Marks
Simplify:
$
\left\{(x+y)^{2 / 3}(x-y)^{3 / 2} \div\left(\sqrt{x+y} \times \sqrt{(x-y)^3}\right)\right\}^6
$
Answer
$\left\{\frac{(x+y)^{2 / 3}(x-y)^{3 / 2}}{\sqrt{x+y} \sqrt{(x-y)^3}}\right\}^6$
$=\left\{\frac{(x+y)^{2 / 3}(x-y)^{3 / 2}}{(x+y)^{1 / 2}(x-y)^{3 / 2}}\right\}^6 \quad\left[\right.$ Using $\left.\sqrt[n]{a}=a^{1 / n}\right]$
$=\left\{(x+y)^{2 / 3} \cdot(x+y)^{-1 / 2} \cdot(x-y)^{3 / 2} \cdot(x-y)^{-3 / 2}\right\}^6$
$=\left\{(x+y)^{\frac{2}{3}-\frac{1}{2}} \cdot(x-y)^{\frac{3}{2}-\frac{3}{2}}\right\}^6 \quad\left[\right.$ Using $\left.a^m \cdot a^n=a^{m+n}\right]$
$=\left\{(x+y)^{\frac{1}{6}}(x-y)^0\right\}^6$
$=\left\{(x+y)^{\frac{1}{6}}\right\}^6 \quad\left[Using\ a^0=1\right]$
$=(x+y)^{\frac{1}{6} \times 6}$$\left[ Using \left(a^m\right)^n=a^{m n}\right]$
$=x+y$
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Question 162 Marks
Find the value of $x$, if $x \sqrt{x}=(x \sqrt{x})^x$.
Answer
Given, or,
$
\begin{aligned}
x \sqrt{x} & =(x \sqrt{x})^x \\
x x^{1 / 2} & =x^x x^{x / 2}
\end{aligned}
$
or,$
\begin{array}{l} 
x^{1+\frac{1}{2}}=x^{x+\frac{x}{2}} \\
{\left[\text { Using } a^m \cdot a^n=a^{m+n}\right] }
\end{array}
$
or,$
x^{\frac{3}{2}}=x^{\frac{3 x}{2}}
$
We know that, if base is equal, then power is also equal i.e.,$
\frac{3}{2}=\frac{3 x}{2}
$
or,$
x=\frac{3}{2} \times \frac{2}{3}=1
$
Hence, value of $x$ is 1 .
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Question 172 Marks
Simplify $2 x^{1 / 2} 3 x^{-1}$ if $x=4$
Answer
We have, $2 x^{1 / 2} 3 x^{-1}$
$\begin{array}{l}=6 \cdot x^{1 / 2} x^{-1} \\ =6 \cdot x^{\frac{1}{2}-1} \quad\left[\text { Using } a^m \cdot a^n=a^{m+n}\right] \\ =6 \cdot x^{-1 / 2} \\ =\frac{6}{x^{1 / 2}} \quad\left[U \operatorname{sing} a^{-n}=\frac{1}{a^n}\right] 1\end{array}$
Now at $x=4$,
$
\frac{6}{(4)^{1 / 2}}=\frac{6}{\left(2^2\right)^{1 / 2}}=\frac{6}{2^{2 \times \frac{1}{2}}}=\frac{6}{2}=3 .
$
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2 Marks Questions - Applied Maths STD 11 Science Questions - Vidyadip