Question 13 Marks
Evaluate $: \frac{\log _5^8}{\left(\log _{25} 16\right) \times\left(\log _{100} 10\right)}$
Answer$ \frac{\log _5 8}{\left(\log _{25} 16\right) \times\left(\log _{100} 10\right)}$
$\Rightarrow \frac{\frac{\log _{10} 8}{\left(\log _{10} 5\right)}}{\frac{\log _{10} 16}{\log _{10} 25} \times \frac{\log _{10} 10}{\log _{10} 100}}$
$\Rightarrow \frac{\frac{\log _{10} 2^3}{\left(\log _{10} 5\right)}}{\frac{\log _{10} 2^4}{\log _{10} 5^2} \times \frac{\log _{10} 10}{\log _{10} 10^2}}$
$\Rightarrow \frac{\log _{10} 2^3}{\log _{10} 5} \times \frac{\log _{10} 5^2}{\log _{10} 2^4} \times \frac{\log _{10} 10^2}{\log _{10} 10}$
$\Rightarrow \frac{3 \log _{10} 2}{\log _{10} 5} \times \frac{2 \log _{10} 5}{4 \log _{10} 2} \times \frac{2 \log _{10} 10}{\log _{10} 10}$
$\Rightarrow 3$
View full question & answer→Question 23 Marks
If $\log _{\sqrt{27}} x=2 \frac{2}{3},$ find $x$
Answer$\log _{\sqrt{27}} x=2 \frac{2}{3}$
$\therefore \log _{\sqrt{27}} x=\frac{8}{3}$
$\therefore x=(\sqrt{27})^{\frac{8}{3}} \ldots\left[\because \log _a x=b \Rightarrow x=a^b\right]$
$ \therefore x=\left(27^{\frac{1}{2}}\right)^{\frac{8}{3}}$
$ \therefore x=\left(3^{\frac{3}{2}}\right)^{\frac{8}{3}}$
$ \therefore x=3^{\left(\frac{3}{2}\right) \times\left(\frac{8}{3}\right)}$
$ \therefore x=3^4$
$ \therefore x=81$
View full question & answer→Question 33 Marks
Show that :$\log_a m \div \log_{ab} m + 1 + \log _ab$
Answer$ \log _{\mathrm{a}} \mathrm{m} \div \log _{\mathrm{ab}} \mathrm{m}==\frac{\log _a m}{\log _{a b} m}$
$ =\frac{\log _m a b}{\log _m a}\left[Q \log _b a=\frac{1}{\log _a b}\right]$
$ =\log _{\mathrm{a}} \mathrm{ab}\left[Q \frac{\log _x a}{\log _{\mathrm{x}} b}=\log _b a\right]$
$ =\log _{\mathrm{a}} \mathrm{a}+\log _{\mathrm{a}} \mathrm{b}$
$ =1+\log _{\mathrm{a}} \mathrm{b}$
View full question & answer→Question 43 Marks
If $\frac{3}{2} \log a+\frac{2}{3} \log b-1=0$, find the value of $a^9 \cdot b^4$.
Answer$ \frac{3}{2} \log a+\frac{2}{3} \log b-1=0$
$ \Rightarrow \log a^{\frac{3}{2}}+\log b^{\frac{2}{3}}=1$
$ \Rightarrow \log \left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right)=1$
$ \Rightarrow \log \left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right)=\log 10$
$ \Rightarrow\left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right)=10$
$ \Rightarrow\left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right)^6=10^6$
$ \Rightarrow a^9 \cdot b^4$
$=10^6$
View full question & answer→Question 53 Marks
Evaluate: $\frac{\log _5 8}{\log _{25} 16 \times \log _{100} 10}$
Answer$ \frac{\log _5 8}{\log _{25} 16 \times \log _{100} 10}$
$ =\frac{\frac{\log _{10} 8}{\log _{10} 5}}{\frac{\log _{10} 16}{\log _{10} 25} \times \frac{\log _{10} 10}{\log _{10} 100}}$
$ =\frac{\frac{\log _{10} 2^3}{\log _{10} 5}}{\frac{\log _{10} 2^4}{\log _{10} 5^2} \times \frac{\log _{10} 10}{\log _{10} 10^2}}$
$ =\frac{\log _{10} 2^3}{\log _{10} 5} \times \frac{\log _{10} 5^2}{\log _{10} 2^4} \times \frac{\log _{10} 10^2}{\log _{10} 10}$
$ =\frac{3 \log _{10} 2}{\log _{10} 5} \times \frac{2 \log _{10} 5}{4 \log _{10} 2} \times \frac{2 \log _{10} 10}{\log _{10} 10}$
$ =3$
View full question & answer→Question 63 Marks
Solve for $\mathbf{x}, \log _x^{15 \sqrt{5 } }=2-\log _x^{3 \sqrt{ 5} }$
Answer$\log_x 15\sqrt 5 = 2 - \log_x3\sqrt 5$
$\Rightarrow \log_x15\sqrt 5 + \log_x3\sqrt 5 = 2$
$\Rightarrow \log_x( 15\sqrt 5 \times 3\sqrt 5 ) = 2$
$\Rightarrow \log_x225 = 2$
$\Rightarrow \log_x15^2 = 2$
$\Rightarrow 2 \log_x15= 2$
$\Rightarrow \log_x15 = 1$
$\Rightarrow x = 15.$
View full question & answer→Question 73 Marks
Given $x = \log_{10}12 , y = \log_4 2 \times \log_{10}9$ and $z = \log_{10}0.4 ,$ find :$(i) x - y - z,(ii) 13^{x - y - z}$
Answer$ \text { (i) } x-y-z$
$ =\log _{10} 12-\log _4 2 \times \log _{10} 9-\log _{10} 0.4$
$ =\log _{10}(4 \times 3)-\log _4 2 x \log _{10} 9-\log _{10} 0.4$
$ =\log _{10} 4+\log _{10} 3-\log _4 2 \times 2 \log _{10} 3-\log _{10}\left(\frac{4}{10}\right)$
$ =\log _{10} 4+\log _{10} 3-\frac{\log _{10} 2}{2 \log _{10} 2} \times 2 \log _{10} 3-\log _{10} 4+\log _{10} 10$
$ =\log _{10} 4+\log _{10} 3-\frac{2 \log _{10} 3}{2}-\log _{10} 4+1$
$=1$
$(ii) 13^{x-y-z}=13^1$
$=13$.
View full question & answer→Question 83 Marks
If $a^2=\log x, b^3=\log y$ and $\frac{a^2}{2}-\frac{b^3}{3}=\log c$, find $c$ in terms of $x$ and $y$.
AnswerGiven $a^2=\log x, b^3=\log y$
Now $\frac{a^2}{2}-\frac{b^3}{3}=\log \mathrm{c}$
$\Rightarrow \frac{\log x}{2}-\frac{\log y}{3}=\log c$
$ \Rightarrow \frac{3 \log x-2 \log y}{6}=\log c$
$ \Rightarrow 3 \log x-2 \log y=6 \log c$
$ \Rightarrow \log x^3-\log y^2=6 \log c$
$ \Rightarrow \log \left(\frac{x^3}{y^2}\right)=\log c^6$
$ \Rightarrow \frac{x^3}{y^2}=c^6$
$ \Rightarrow \mathrm{c}=\sqrt[6]{\frac{x^3}{y^2}}$
View full question & answer→Question 93 Marks
Solve for $\mathbf{x}$, if $: \log _x 49-\log _x 7+\log _x \frac{1}{343}+2=0$
Answer$ \log _x 49-\log _x 7+\log _x \frac{1}{343}=-2$
$ \Rightarrow \log _x \frac{49}{7 \times 343}=-2$
$ \Rightarrow \log _x \frac{1}{49}=-2$
$ \Rightarrow-\log _x 49=-2$
$ \Rightarrow \log _x 49=2$
$\Rightarrow 49=x^2 \ldots [$Removing logarithm$]$
$\therefore x=7$
View full question & answer→Question 103 Marks
Solve: $\log_5( x + 1 ) - 1 = 1 + \log_5( x - 1 ).$
Answer$ \log _5(x+1)-1=1+\log _5(x-1)$
$ \Rightarrow \log _5(x+1)-\log _5(x-1)=2$
$ \Rightarrow \log _5 \frac{x+1}{x-1}=2$
$ \Rightarrow \frac{x+1}{x-1}=5^2$
$ \Rightarrow \frac{x+1}{x-1}=25$
$ \Rightarrow x+1=25(x-1)$
$ \Rightarrow x+1=25 x-25$
$ \Rightarrow 25 x-x=25+1$
$ \Rightarrow 24 x=26$
$ \Rightarrow x=\frac{26}{24}$
$=\frac{13}{12}$
View full question & answer→Question 113 Marks
Given $\log _{10} \mathrm{x}=2 \mathrm{a}$ and $\log _{10} \mathrm{y}=\frac{b}{2}$. If $\log _{10}^p=3 a-2 b$, express $\mathrm{P}$ in terms of $x$ and $y$.
AnswerWe know $10^a=x^{1 / 2}$
$10^{\mathrm{b} / 2}=\mathrm{y}$
$ \Rightarrow 10^{\mathrm{b}}=\mathrm{y}^2$
$\log _{10}^p=3 a-2 b$
$ \Rightarrow p=10^{3 a}-2 b$
$ \Rightarrow p=\left(10^3\right)^a \div\left(10^2\right)^b$
$ \Rightarrow p=(10 a)^3 \div\left(10^b\right)^2$
Substituting $10^{\mathrm{a}} \ 10^{\mathrm{b}}$, We get
$\Rightarrow \mathrm{p}=\left(\mathrm{x}^{1 / 2}\right)^3 \div\left(\mathrm{y}^2\right)^2$
$\Rightarrow \mathrm{p}=x^{\frac{3}{2}} \div y^4$
$\Rightarrow \mathrm{p}=\frac{x^{\frac{3}{2}}}{y^4}$
View full question & answer→Question 123 Marks
If $p = \log 20$ and $q = \log 25 ,$ find the value of $x ,$ if $2\log( x + 1 ) = 2p - q.$
AnswerGiven that
$p=\log 20$ and $q=\log 25$
We also have
$2 \log (x+1)=2 p-q$
$ \Rightarrow 2 \log (x+1)=2 \log 20-\log 25$
$ \Rightarrow \log (x+1)^2=\log 20^2-\log 25$
$ \Rightarrow \log (x+1)^2=\log 400-\log 25$
$ \Rightarrow \log (x+1)^2=\log \frac{400}{25}$
$ \Rightarrow \log (x+1)^2=\log 16$
$ \Rightarrow \log (x+1)^2=\log 4^2$
$ \Rightarrow x+1=4$
$ \Rightarrow x=4-1$
$ \Rightarrow x=3 .$
View full question & answer→Question 133 Marks
If $2 \log y - \log x - 3 = 0,$ express $x$ in terms of $y.$
Answer$ 2 \log y-\log x-3=0$
$ \Rightarrow 2 \log y-\log x=3$
$ \Rightarrow \log y^2-\log x=3$
$ \Rightarrow \log y^2-\log x=\log 1000$
$ \Rightarrow \log \frac{y^2}{x}=\log 1000$
$ \Rightarrow \frac{y^2}{x}=1000$
$ \Rightarrow x=\frac{y^2}{1000}$
View full question & answer→Question 143 Marks
If $\log (a + 1) = \log (4a - 3) - \log 3;$ find $a.$
AnswerGiven that
$\log (a+1)=\log (4 a-3)-\log 3$
$ \Rightarrow \log (a+1)=\log \left(\frac{4 a-3}{3}\right)$
$ \Rightarrow a+1=\frac{4 a-3}{3}$
$ \Rightarrow 3 a+3=4 a-3$
$ \Rightarrow 4 a-3 a=3+3$
$ \Rightarrow a=6$
View full question & answer→Question 153 Marks
Prove that : If $a \log b + b \log a - 1 = 0,$ then $b^a.a^b= 10$
AnswerGiven that $\log b + b \log a - 1 = 0$
$\Rightarrow a \log b + b \log a = 1$
$\Rightarrow \log b^a + loga^b =1$
$\Rightarrow \log b^a + \log a^b = \log 10$
$\Rightarrow \log ( b^a . a^b ) = \log 10$
$\Rightarrow b^a . a^b = 10$
View full question & answer→Question 163 Marks
Prove that: $(\log a)^2-(\log b)^2=\log \left(\frac{a}{b}\right) \cdot \log (a b)$
Answer$\text { L.H.S }=(\log a)^2-(\log b)^2$
$\Rightarrow \text { L.H.S }=(\log a+\log b)(\log a-\log b)$
$\Rightarrow \text { L.H.S }=\log (a b) \log \left(\frac{a}{b}\right)$
$\Rightarrow \text { L.H.S }=\log \left(\frac{a}{b}\right) \times \log (a b)$
$\Rightarrow \text { L.H.S }=\text { R.H.S }$
Hence proved.
View full question & answer→Question 173 Marks
If $\log (a + b) = \log a + \log b,$ find $a$ in terms of $b.$
Answer$ \log (a+b)=\log a+\log b$
$ \Rightarrow \log (a+b)=\log a b$
$ \Rightarrow a+b=a b$
$ \Rightarrow a-a b=-b$
$ \Rightarrow-a b+a=-b$
$ \Rightarrow-a(b-1)=-b$
$ \Rightarrow a(b-1)=b$
$ \Rightarrow a=\frac{b}{b-1}$
View full question & answer→Question 183 Marks
Given: $\log_3m = x$ and $\log_3n = y.$If $2 \log_3A = 5x - 3y;$ find $A$ in terms of $m$ and $n.$
AnswerGiven that $\log _3 m=x$ and $\log _3 n=y$
$\Rightarrow 3^{\mathrm{x}}=\mathrm{m} \text { and } 3^{\mathrm{y}}=\mathrm{n}$
Consider the given expression :
$2 \log _3 A=5 x-3 y$
$ \Rightarrow 2 \log _3 A=5 \log _3 m-3 \log _3 n$
$ \Rightarrow \log _3 A^2=\log _3 m^5-\log _3 n^3$
$ \Rightarrow \log _3 A^2=\log _3\left(\frac{m^5}{n^3}\right)$
$ \Rightarrow A^2=\left(\frac{m^5}{n^3}\right)$
$ \Rightarrow A=\sqrt{\left(\frac{m^5}{n^3}\right)}$
View full question & answer→Question 193 Marks
Given: $\log_3m = x$ and $\log_3n = y.$Write down $3^{1 - 2y +3x}$ in terms of $m$ and $n.$
AnswerGiven that $\log _3 m=x$ and $\log _3 n=y$
$\Rightarrow 3^{\mathrm{x}}=\mathrm{m}$ and $3^{\mathrm{y}}=\mathrm{n}$
Consider the given expression :
$3^{1-2 y+3 x}=3^1 \cdot 3^{-2 y} \cdot 3^{3 x}$
$ =3 \cdot \frac{1}{3^{2 y}} \cdot 3^{3 x}$
$ =\frac{3}{\left(3^y\right)^2} \cdot\left(3^x\right)^3$
$ =\frac{3}{(n)^2} \cdot(m)^3$
$ =\frac{(3 m)^3}{n^2}$
Therefore, $3^{1-2 y+3 x}=\frac{(3 m)^3}{n^2}$
View full question & answer→Question 203 Marks
Given: $\log_3m = x$ and $\log_3n = y.$Express $3^{2x - 3}$ in terms of $m.$
AnswerGiven that $\log _3 m=x$ and $\log _3 n=y$
$\Rightarrow 3^{\mathrm{x}}=\mathrm{m}$ and $3^{\mathrm{y}}=\mathrm{n}$
Consider the given expression :
$3^{2 x-3}$
$ =3^{2 x} \cdot 3^{-3}$
$ =3^{2 x} \cdot \frac{1}{3^3}$
$=\frac{3^{2 x}}{3^3}$
$=\frac{\left(3^x\right)^2}{3^3}$
$=\frac{m^2}{27}$
Therefore, $3^{2 x-3}=\frac{m^2}{27}$
View full question & answer→Question 213 Marks
If $\log 27 = 1.431$, find the value of $: \log 300$
Answer$ \log 27=1.431$
$ \Rightarrow \log 3 \times 3 \times 3=1.431$
$ \Rightarrow \log 3^3=1.431$
$ \Rightarrow 3 \log 3=1.431$
$ \Rightarrow \log 3=\frac{1.431}{3}$
$\Rightarrow \log 3=0.477$
$\ldots(1)$
$ \log 300$
$ =\log (3 \times 100)$
$ =\log 3+\log 100$
$ =\log 3+2 \ldots\left[\because \log _{10} 100=2\right]$
$ =0.477+2$
$ =2.477$
View full question & answer→Question 223 Marks
If $\log 27 = 1.431,$ find the value of :$ \log 9$
Answer$ \log 27=1.431$
$ \Rightarrow \log 3 \times 3 \times 3=1.431$
$ \Rightarrow \log 3^3=1.431$
$ \Rightarrow 3 \log 3=1.431$
$ \Rightarrow \log 3=\frac{1.431}{3}$
$\Rightarrow \log 3=0.477\ldots(1)$
$ \log 9$
$ =\log (3 \times 3)$
$ =\log 32$
$ =2 \log 3$
$ =2 \times 0.477$
$ =0.954$
View full question & answer→Question 233 Marks
Given $\log x=2 m-n, \log y=n-2 m$ and $\log z=3 m-2 n$, find in terms of $m$ and $n$, the value of $\log \frac{x^2 y^3}{z^4}$.
AnswerGiven $\log x=2 m-n, \log y=n-2 m, \log z=3 m-2 n$.
Given : $\log \frac{x^2 y^3}{z^4}$
We know that $\log (a / b)=\log a-\log b$.
$\Rightarrow \log \left(x^2 y^3\right)-\log \left(z^4\right)$
We know that $\log (a b)=\log a+\log b$
$\Rightarrow \log \left(x^2\right)+\log \left(y^3\right)-\log \left(z^4\right)$
$ \Rightarrow 2 \log x+3 \log y-4 \log z$
$ \Rightarrow 2(2 m-n)+3(n-2 m)-4(3 m-2 n)$
$ \Rightarrow 4 m-2 n+3 n-6 m-12 m+8 n$
$ \Rightarrow-14 m+9 n$
View full question & answer→Question 243 Marks
Given that $\log x=m+n$ and $\log y=m-n,$ express the value of $\log\frac{10 x}{y^2}$ in terms of $\mathrm{m}$ and $\mathrm{n}$.
AnswerGiven that
$\log x=m+n ;$
$ \log y=m-n ;$
Consider the expression $\log \frac{10 x}{y^2}$ :
$ \log \frac{10 x}{y^2}$
$ =\log 10 x-\log y^2$
$ \Rightarrow \log 10 x-2 \log y \ldots\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow \log 10+\log x-2 \log y \ldots\left[\log _a m+\log _a n=\log _a m n\right]$
$ \Rightarrow 1+\log x-2 \log y$
$ \Rightarrow 1+m+n-2(m-n)$
$ \Rightarrow 1+m+n-2 m+2 n$
$ \Rightarrow \log \frac{10 x}{y^2}$
$=1-m+3 n .$
View full question & answer→Question 253 Marks
Solve for $x :\frac{\log 225}{\log 15}=\log x$
Answer$ \Rightarrow \log x=\frac{\log 225}{\log 15}$
$ \Rightarrow \log x=\frac{\log 15 \times 15}{\log 15}$
$ \Rightarrow \log x=\frac{\log 15^2}{\log 15}$
$ \Rightarrow \log x=\frac{2 \log 15}{\log 15} \ldots\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow \log x=2$
$ \Rightarrow \log _{10} x=2$
$ \Rightarrow 10^2=x$
$ \Rightarrow x=10 \times 10$
$ \Rightarrow x=100$
View full question & answer→Question 263 Marks
Solve for $\mathbf{x}: \frac{\log 128}{\log 32}=x$
Answer$ \frac{\log 128}{\log 32}=\mathrm{x}$
$ \Rightarrow x=\frac{\log 128}{\log 32}$
$ \Rightarrow x=\frac{\log 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{\log 2 \times 2 \times 2 \times 2 \times 2}$
$ \Rightarrow x=\frac{\log 2^7}{\log 2^5}$
$ \Rightarrow x=\frac{7 \log 2}{5 \log 2} \ldots\left[\mathrm{n} \log _{\mathrm{a}} \mathrm{m}=\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}\right]$
$ \Rightarrow x=\frac{7}{5}$
$ \Rightarrow \mathrm{x}=1.4$
View full question & answer→Question 273 Marks
Solve for $\mathbf{x}: \frac{\log 81}{\log 27}=x$
Answer$ \frac{\log 81}{\log 27}=\mathrm{x}$
$ \Rightarrow \mathrm{x}=\frac{\log 81}{\log 27}$
$ \Rightarrow \mathrm{x}=\frac{\log 3 \times 3 \times 3 \times 3}{\log 3 \times 3 \times 3}$
$ \Rightarrow \mathrm{x}=\frac{\log 3^4}{\log 3^3}$
$ \Rightarrow \mathrm{x}=\frac{4 \log 3}{3 \log 3} \ldots\left[\mathrm{n} \log _{\mathrm{a}} \mathrm{m}=\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}\right]$
$ \Rightarrow \mathrm{x}=\frac{4}{3}$
$ \Rightarrow \mathrm{x}=1 \frac{1}{3}$
View full question & answer→Question 283 Marks
Solve for $x$ :$\log (x - 2) + \log (x + 2) = \log 5$
Answer$\log (x-2)+\log (x+2)=\log 5$
$\Rightarrow \log (x-2)(x+2)=\log 5 \ldots\left[\log _a m+\log _a n=\log _a m n\right]$
$ \Rightarrow \log \left(x^2-4\right)=\log 5$
$ \Rightarrow x^2-4=5$
$ \Rightarrow x^2=9$
$ \Rightarrow x= \pm \sqrt{9}$
$ \Rightarrow x= \pm \sqrt{3^2}$
$ \Rightarrow x= \pm 3$
View full question & answer→Question 293 Marks
Solve for $x : \log (x^2 - 21) = 2.$
Answer$ \log \left(x^2-21\right)=2$
$ \Rightarrow \log \left(x^2-21\right)=\log 100$
$ \Rightarrow x^2-21-100=0$
$ \Rightarrow x^2-121=0$
$ \Rightarrow x^2=121$
$ \Rightarrow x= \pm \sqrt{121}$
$ \Rightarrow x= \pm 11$
View full question & answer→Question 303 Marks
Evaluate the following without using tables :$\log 4+\frac{1}{3} \log 125-\frac{1}{5} \log 32$
AnswerConsider the given expression
$\log 4+\frac{1}{3} \log 125-\frac{1}{5} \log 32$
$ =\log 4+\log (125)^{\frac{1}{3}}-\log (32)^{\frac{1}{5}} \ldots\left[n \log _a m=\log _a m^n\right]$
$ =\log 4+\log \left(5^3\right)^{\frac{1}{3}}-\log \left(2^5\right)^{\frac{1}{5}}$
$ =\log 4+\log 5-\log 2$
$ =\log 4 \times 5-\log 2 \ldots .\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ =\log \left(\frac{20}{2}\right) \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log 10$
$ =1$
View full question & answer→Question 313 Marks
Evaluate the following without using tables $:\log 5 + \log 8 - 2 \log 2$
AnswerConsider the given expression
$\log 5+\log 8-2 \log 2 \ldots . .\left[n \log _a m=\log _a m^n\right]$
$ =\log 5+\log 8 \times 8-\log 2^2 \ldots . .\left[n \log _a m=\log _a m^n\right]$
$ =\log 5 \times 8-\log 2^2$
$ =\log 40-\log 4$
$=\log \frac{40}{4} \ldots .\left[\log _{\mathrm{a}} \mathrm{m}-\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}}\left(\frac{m}{n}\right)\right]$
$ =\log 10$
$ =1$
View full question & answer→Question 323 Marks
Given $3 \log x+\frac{1}{2} \log y=2$, express $y$ in term of $x$.
Answer$3 \log x+\frac{1}{2} \log y=2$
$ \Rightarrow \log x^3+\log \sqrt{y}=2$
$ \Rightarrow \log x^3 \sqrt{y}=2$
$ \Rightarrow x^3 \sqrt{y}=10^2$
$ \Rightarrow \sqrt{y}=\frac{10^2}{x^3}$
Squaring both sides, we get
$y=\frac{10000}{x^6}$
$\Rightarrow y=10000 x^{-6}$
View full question & answer→Question 333 Marks
Express in terms of $\log 2$ and $\log 3 :\log \frac{26}{51}-\log \frac{91}{119}$
Answer$ \log \frac{26}{51}-\log \frac{91}{119}$
$ =\log \left(\frac{\frac{26}{51}}{\frac{91}{119}}\right) \ldots\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log \frac{26}{51} \times \frac{119}{91}$
$ =\log \frac{2 \times 13}{3 \times 17} \times \frac{7 \times 17}{7 \times 13}$
$ =\log \frac{2}{3}$
$ =\log 2-\log 3 \ldots \ldots .\left[\log _{\mathrm{a}} \frac{m}{n}=\log _{\mathrm{a}} \mathrm{m}-\log _{\mathrm{a}} \mathrm{n}\right]$
View full question & answer→Question 343 Marks
Express in terms of $\log 2$ and $\log 3$ : $\log 4.5$
Answer$ \log 4.5$
$ =\log \frac{45}{10}$
$ =\log \frac{5 \times 3 \times 3}{5 \times 2}$
$ =\log \frac{3^2}{2}$
$ =\log 3^2-\log 2 \ldots . .\left[\log _{\mathrm{a}} \frac{m}{n}=\log _{\mathrm{a}} \mathrm{m}-\log _{\mathrm{a}} \mathrm{n}\right]$
$ =2 \log 3-\log 2 \ldots \ldots .\left[\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}=\mathrm{n} \log _{\mathrm{a}} \mathrm{m}\right]$
View full question & answer→Question 353 Marks
If $\log 2=0.3010$ and $\log 3=0.4771 ;$ find the value of $: \frac{2}{3} \log 8$
AnswerWe know that $\log 2=0.3010$ and $\log 3=0.4771$
$\frac{2}{3} \log 8$
$ =\frac{2}{3} \log 2 \times 2 \times 2$
$ =\frac{2}{3} \log 2^3$
$ =3 \times \frac{2}{3} \log 2 \ldots\left[\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}=\mathrm{n} \log _{\mathrm{a}} \mathrm{m}\right]$
$ =2 \log 2$
$ =2 \times 0.3010 \ldots[\because \log 2=0.3010]$
$ =0.602$
View full question & answer→Question 363 Marks
If$ \log 2 = 0.3010$ and $\log 3 = 0.4771;$ find the value of $: \log 25$
AnswerWe know that $\log 2=0.3010$ and $\log 3=0.4771$
$\log 25$
$=\log \left(\frac{25}{4} \times 4\right)$
$ =\log \left(\frac{100}{4}\right) \ldots\left[\log _a m n=\log _a m+\log _a n\right]$
$ =\log 100-\log (2 \times 2) \ldots\left[\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n\right]$
$=2-\log \left(2^2\right) \ldots[\log 100=2]$
$=2-2 \log 2 \ldots\left[\log _a m^n=n \log _a m\right]$
$=2-2(0.3010) \ldots[\because \log 2=0.3010]$
$=1.398 $
View full question & answer→Question 373 Marks
If $\log 2 = 0.3010$ and $\log 3 = 0.4771;$ find the value of $: \log 15$
AnswerWe know that $\log 2=0.3010$ and $\log 3=0.4771. \log 15$
$=\log \left(\frac{15}{10} \times 10\right)$
$=\log \left(\frac{15}{10}\right)+\log 10$
$=\log \left(\frac{3}{2}\right)+1 \ldots[\because \log 10=1]$
$ =\log 3-\log 2+1 \ldots\left[\because \log m-\log n=\log \left(\frac{m}{n}\right)\right]$
$ =0.4771-0.3010+1$
$ =1.1761$
View full question & answer→Question 383 Marks
If $\log 2 = 0.3010$ and $\log 3 = 0.4771 ;$ find the value of : $\log 12$
AnswerWe know that $\log 2 = 0.3010$ and $\log 3 = 0.4771$
$\log 12$
$= \log 2 \times 2 \times 3$
$= \log 2 \times 2 + \log 3 ...[ \log_amn = \log_am + \log_an ]$
$= \log2^2 + log3$
$= 2\log2 + \log 3 ...[n \log_am = \log_am^n]$
$= 2( 0.3010 ) + 0.4771 ...[ \because \log 2 = 0.3010$ and $\log3 = 0.4771 ]$
$= 1.0791$
View full question & answer→Question 393 Marks
If $\log _{10} 2=a$ and $\log _{10} 3=b ;$ express each of the following in terms of $'a\ '$ and $'b\ ' :\log 3 \frac{1}{8}$
Answer$ \log 3 \frac{1}{8}$
$ =\log _{10}\left(\frac{25}{8} \times \frac{4}{4}\right)$
$ =\log _{10}\left(\frac{100}{32}\right)$
$ =\log _{10} 100-\log _{10} 32 \ldots\left[\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n\right]$
$ =\log _{10} 100-\log _{10} 2^5$
$=2-\log _{10} 2^5 \ldots\left[\because \log _{10} 100=2\right]$
$=2-5 \log _{10} 2 \ldots\left[\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}=\mathrm{n} \log _{\mathrm{a}} \mathrm{m}\right]$
$=2-5 \mathrm{a} \ldots\left[\because \log _{10} 2=\mathrm{a}\right]$
View full question & answer→Question 403 Marks
If $log_{10}2 = a$ and $log_{10}3 = b ;$ express each of the following in terms of $'a\ '$ and $'b\ ' : \log 5.4$
AnswerGiven that $\log _{10} 2=a$ and $\log _{10} 3=b$
$\log 5.4$
$ =\log \frac{54}{10}$
$=\log \frac{2 \times 3 \times 3 \times 3}{10}$
$ =\log (2 \times 3 \times 3 \times 3)-\log _{10} 10 \ldots\left[\log _a m=\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log _{10} 2+\log _{10} 3^3-\log _{10} 10 \ldots\left[\log _a m n=\log _a m+\log _a n\right]$
$ =\log _{10} 2+3 \log _{10} 3-\log _{10} 10 \ldots\left[n \log _a m=\log _a m^n\right]$
$ =\log _{10} 2+3 \log _{10} 3-1 \ldots\left[\because \log _{10} 10=1\right]$
$ =a+3 b-1 \ldots\left[\because \log _{10} 2=a \text { and } \log _{10} 3=b\right]$
View full question & answer→Question 413 Marks
If $\log _{10} 2=a$ and $\log _{10} 3=b;$ express each of the following in terms of $'a\ '$ and$ 'b\ ': \log 2 \frac{1}{4}$
AnswerGiven that $\log _{10} 2=a$ and $\log _{10} 3=b$
$
\log 2 \frac{1}{4}$
$ =\log \left(\frac{9}{4}\right)$
$ =\log \left(\frac{3}{2}\right)^2$
$=2 \log \left(\frac{3}{2}\right) \ldots\left[\operatorname{nog}_{\mathrm{a}} \mathrm{m}=\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}\right]$
$=2(\log 3-\log 2) \ldots\left[\log _{\mathrm{a}} \mathrm{m}-\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \frac{m}{n}\right]$
$=2(\mathrm{~b}-\mathrm{a}) \ldots\left[\left[\because \log _{10} 2=\mathrm{a} \text { and } \log _{10} 3=\mathrm{b}\right]\right.$
$=2 \mathrm{~b}-2 \mathrm{a} $
View full question & answer→Question 423 Marks
If $\log _{10} 2=a$ and $\log _{10} 3=b;$ express each of the following in terms of $'a\ '$ and $'b\ ': \log 12$
AnswerGiven that $\log_{10}2 = a$ and $\log_{10}3 = b$
$\log 12$
$= \log 2 \times 2 \times 3$
$= \log 2 \times 2 + \log 3 ...[ \log_amn = \log_am + \log_an ]$
$= \log 2^2 + \log 3$
$= 2\log 2 + \log 3 ...[ n \log_am = \log_am^n ]$
$= 2a + b ...[ \because \log_{10}2 = a$ and $ \log_{10}3 = b ]$
View full question & answer→Question 433 Marks
Evaluate $:\log_{0.5}16$
AnswerLet $\log _{0.5} 16=x$
$\Rightarrow 0.5^x=16$
$ \Rightarrow\left(\frac{5}{10}\right)^x=2 \times 2 \times 2 \times 2$
$\Rightarrow\left(\frac{1}{2}\right)^x=2^4$
$\Rightarrow \frac{1}{2^x}=2^4$
$\Rightarrow 2^{-x}=2^4$
$ \Rightarrow-x=4$
$ \Rightarrow \mathrm{x}=-4$
Thus, $\log _{0.5} 16=-4$
View full question & answer→Question 443 Marks
Evaluate : $\log_{16}8$
Answer$\text { Let } \log _{16} 8=x$
$ \Rightarrow 16^x=8$
$ \Rightarrow(2 \times 2 \times 2 \times 2)^x=2 \times 2 \times 2$
$ \Rightarrow\left(2^4\right)^x=2^3$
$ \Rightarrow 2^{4 x}=2^3$
$ \Rightarrow 4 x=3$
$ \Rightarrow x=\frac{3}{4}$
Thus, $\log _{16} 8=\frac{3}{4}$
View full question & answer→Question 453 Marks
Evaluate $:\log_2 ( 1\div 8 )$
AnswerLet $\log _2 \frac{1}{8}=x$
$\Rightarrow 2^x=\frac{1}{8}$
$ \Rightarrow 2^x=\frac{1}{2 \times 2 \times 2}$
$ \Rightarrow 2^x=\frac{1}{2^3}$
$ \Rightarrow 2^x=2^{-3}$
$ \Rightarrow x=-3$
Thus, $\log _2 \frac{1}{8}=-3$
View full question & answer→Question 463 Marks
Find $x,$ if $:\log_7(2x^2- 1) = 2$
AnswerConsider the equation
$ \log _7\left(2 x^2-1\right)=2 $
$ \Rightarrow 7^2=2 x^2-1 $
$ \Rightarrow 7 x 7=2 x^2-1 $
$ \Rightarrow 2 x^2-1-49=0 $
$ \Rightarrow 2 x^2-50=0 $
$ \Rightarrow 2 x^2=50 $
$ \Rightarrow x^2=\frac{50}{2} $
$ \Rightarrow x^2=25 $
$ \Rightarrow x= \pm \sqrt{25}$
$\Rightarrow x=5 .....[ $neglecting the negative value$ ]$
View full question & answer→Question 473 Marks
Find $x,$ if $:\log_432 = x - 4$
AnswerConsider the equation
$ \log _4 32=x-4 $
$ \Rightarrow 4^{x-4}=32 $
$ \Rightarrow\left(2^2\right)^{x-4}=2^5 $
$ \Rightarrow 2^{2(x-4)}=2^5 $
$ \Rightarrow 2 x-8=5 $
$ \Rightarrow 2 x=5+8 $
$ \Rightarrow 2 x=13$
$\Rightarrow x=\frac{13}{2}$
$\Rightarrow x=6 \frac{1}{2}$
View full question & answer→Question 483 Marks
Find $x,$ if : $\log_9243 = x$
AnswerConsider the equation
$ \log _9 243=x$
$ \Rightarrow 9^x=243 $
$ \Rightarrow\left(3^2\right)^x=3^5$
$\Rightarrow 3^{2 x}=3^5$
$\Rightarrow 2 x=5$
$\Rightarrow x=\frac{5}{2}$
$\Rightarrow x=2 \frac{1}{2}$
View full question & answer→Question 493 Marks
Find the logarithm of$ : 27$ to the base $9$
Answer Let $\log _9 27=\mathrm{x}$
$ \therefore 9^{\mathrm{x}}=27$
$ \Rightarrow(3 \times 3)^{\mathrm{x}}=3 \times 3 \times 3$
$ \Rightarrow\left(3^2\right)^{\mathrm{x}}=3^3$
$ \Rightarrow\left(3^2\right)^{\mathrm{x}}=3^3$
$ \Rightarrow 2 x=3 \ldots . .\left[\right.$ If $a^m=a^n$; then $\left.m=n\right]$
$ \Rightarrow x=\frac{3}{2}$
$ \therefore \log _9 27=\frac{3}{2}$
View full question & answer→Question 503 Marks
Find the logarithm of : $\frac{1}{16}$ to the base $4$
AnswerLet $\log _4 \frac{1}{16}=x$
$\therefore 4^x=\frac{1}{16}$
$\Rightarrow 4^x=\frac{1}{4 \times 4}$
$\Rightarrow 4^x=(4 \times 4)^{-1}$
$\Rightarrow 4^x=\left(4^2\right)^{-1}$
$\Rightarrow 4^x=4^{-2}$
$\Rightarrow x=-2 \ldots .\left[\right.$ If $a^m=a^n$; then $\left.m=n\right]$
$\therefore \log _4 \frac{1}{16}=-2$
View full question & answer→