Question 14 Marks
$\log _2 x+\log _4 x+\log _{16} x=\frac{21}{4}$
Answer$\log _2 x+\log _4 x+\log _{16} x=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}+\frac{1}{\log _x 2^2}+\frac{1}{\log _x 2^4}=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}+\frac{1}{2 \log _x 2}+\frac{1}{4 \log _x 2}=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}\left(1+\frac{1}{2}+\frac{1}{4}\right)=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}\left(\frac{7}{4}\right)=\frac{21}{4} $
$\therefore \log _x 2=\frac{7}{4} \cdot \frac{4}{21}$
$ \therefore \log _x 2=\frac{1}{3}$
$\therefore x^{\frac{1}{3}}=2$
$ \therefore x=2^3 $
$ =8 .$
View full question & answer→Question 24 Marks
Solve the following:$\log _4 x+\log _4(x-6)=2$
Answer$\log _4 x + \log_4(x-6) = 2$
$\Rightarrow \log _4{x(x-6)} = 2 \log _4 4$
$\Rightarrow \log _4 {x^2- 6x} = \log _4 4^2$
$\Rightarrow x^2 - 6x = 16$
$\Rightarrow x^2 - 6x - 16 = 0$
$\Rightarrow x^2 - 8x + 2x - 16 = 0$
$\Rightarrow x (x - 8) + 2 ( x - 8) = 0$
$\Rightarrow (x - 8)(x + 2) = 0$
$\Rightarrow x = 8$ or $-2$
Negative value is rejected
So, $x = 8$.
View full question & answer→Question 34 Marks
Simplify the following:$12 \log \frac{3}{2}+7 \log \frac{125}{27}-5 \log \frac{25}{36}-7 \log 25+\log \frac{16}{3}$
Answer$12 \log \frac{3}{2}+7 \log \frac{125}{27}-5 \log \frac{25}{36}-7 \log 25+\log \frac{16}{3}$
$=12 \log \frac{3}{2}+7 \log \frac{5^3}{3^3}-5 \log \frac{5^2}{2^2 \times 3^2}-7 \log 5^2+\log \frac{2^2}{3} $
$ =12 \log 3-12 \log 2+7 \log 5^3-7 \log 3^3-5 \log 5^2+5 \log 2^2+ $
$5 \log 3^2-7 \log 5^2+\log 2^4-\log 3$
$ =12 \log 3-12 \log 2+21 \log 5-21 \log 3-10 \log 5+10 \log 2 $
$+10 \log 3-14 \log 5+4 \log 2-\log 3$
$=2 \log 2+2 \log 5 .$
View full question & answer→Question 44 Marks
Simplify the following:$2 \log 7+3 \log 5-\log \frac{49}{8}$
Answer$2 \log 7+3 \log 5-\log \frac{49}{8}$
$ =2 \log 7+3 \log 5-\log 49+\log 8$
$ =2 \log 7+3 \log 5-\log 7^2+\log 2^3 $
$ =2 \log 7+3 \log 5-2 \log 7+3 \log 2 $
$ =3 \log 5+3 \log 2 $
$=3(\log 5+\log 2)$
$ =3 \log (5 \times 2) $
$ =3 \log 10$
$=3 \times 1 $
$ =3 .$
View full question & answer→Question 54 Marks
Simplify the following:$2 \log 5+\log 8-\frac{1}{2} \log 4$
Answer$2 \log 5+\log 8-\frac{1}{2} \log 4$
$ =2 \log 5+\log 2^3-\frac{1}{2} \log 2^2$
$ =2 \log 5+3 \log 2-\frac{1}{2} \times 2 \log 2$
$ =2 \log 5+3 \log 2-\log 2$
$ =2 \log 5+2 \log 2 $
$ =2(\log 5+\log 2) $
$=2 \log (5 \times 2)$
$ =2 \log 10 $
$ =2 \times 1 $
$ =2 .$
View full question & answer→Question 64 Marks
Express the following as a single logarithm:$3 \log \frac{5}{8}+2 \log \frac{8}{15}-\frac{1}{2} \log \frac{25}{81}+3$
Answer$3 \log \frac{5}{8}+2 \log \frac{8}{15}-\frac{1}{2} \log \frac{25}{81}+3 $
$=3 \log \frac{5}{2^3}+2 \log \frac{2^3}{3 \times 5}-\frac{1}{2} \log \frac{5^2}{3^4}+3 \log 10 $
$ =3 \log 5-3 \log 2^3+2 \log 2^3-2 \log 3-2 \log 5-\frac{1}{2} \log 5^2+\frac{1}{2} \log 3^4+3 \log (2 \times 5)$
$=3 \log 5-3 \times 3 \log 2+2 \times 3 \log 2-2 \log 3-2 \log 5-\frac{1}{2} \times 2 \log 5+\frac{1}{2} \times 4 \log 3+3 \log 2+3 \log 5$
$ =3 \log 5-9 \log 2+6 \log 2-2 \log 3-2 \log 5-\log 5+2 \log 3+3 \log 2+3 \log 5 $
$=3 \log 5 $
$ =\log 5^3 $
$ =\log 125 .$
View full question & answer→Question 74 Marks
Express the following as a single logarithm:$\frac{1}{2} \log 25-2 \log 3+\log 36$
Answer$\frac{1}{2} \log 25-2 \log 3+\log 36 $
$ =\frac{1}{2} \log 5^2-2 \log 3+\log \left(2^2 \times 3^2\right) $
$ =\frac{1}{2} \times 2 \log 5-2 \log 3+\log 2^2+\log 3^2$
$ =\log 5+2 \log 2 $
$ =\log 5+\log 2^2 $
$ =\log 5+\log 4$
$=\log (5 \times 4) $
$ =\log 20 .$
View full question & answer→Question 84 Marks
Express the following as a single logarithm:$2 \log \frac{16}{25}-3 \log \frac{8}{5}+\log 90$
Answer$2 \log \frac{16}{25}-3 \log \frac{8}{5}+\log 90$
$=2 \log \frac{2^4}{5^2}-3 \log \frac{2^3}{5}+\log \left(2 \times 5 \times 3^2\right)$
$=2 \log 2^4-2 \log 5^2-3\left\{\log 2^3-\log 5\right\}+\log 2+\log 5+\log 3^2$
$=2 \times 4 \log 2-2 \times 2 \log 5-3 \times 3 \log 2+3 \log 5+\log 2 \log 5+$
$2 \log 3$
$=8 \log 2-4 \log 5-9 \log 2+3 \log 5+\log 2+\log 5+2 \log 3$
$=2 \log 3$
$=\log 3^2$
$=\log 9 .$
View full question & answer→Question 94 Marks
Express the following as a single logarithm:$2 \log \frac{9}{5}-3 \log \frac{3}{5}+\log \frac{16}{20}$
Answer$2 \log \frac{9}{5}-3 \log \frac{3}{5}+\log \frac{16}{20}$
$=2 \log 9-2 \log 5-3 \log 3+3 \log 5+\log 16-\log 20$
$=2 \log \left(3^2\right)-2 \log 5-3 \log 3+3 \log 5+\log \left(4^2\right)-\log (5 \times 4)$
$=4 \log 3-2 \log 5-3 \log 3+3 \log 5+2 \log 4-\log 5-\log 4$
$=(4-3) \log 3+(-2-1+3) \log 5+\log 4$
$=\log 3+\log 4$
$=\log (3 \times 4)$
$=\log 12 .$
View full question & answer→Question 104 Marks
Express the following as a single logarithm:$2+\frac{1}{2} \log 9-2 \log 5$
Answer$2+\frac{1}{2} \log 9-2 \log 5$
$=2+\frac{1}{2} \log 3^2-2 \log 5 $
$=2 \log 10+\frac{1}{2} \times 2 \log 3-2 \log 5 $
$=\log 10^2+\log 3-\log 5^2$
$ =\log 100+\log 3-\log 25 $
$=\log \frac{100 \times 3}{25} $
$ =\log 12 .$
View full question & answer→Question 114 Marks
Express the following as a single logarithm:$2 \log 3-\frac{1}{2} \log 16+\log 12$
Answer$2 \log 3-\frac{1}{2} \log 16+\log 12$
$=2 \log 3-\frac{1}{2} \log 2^4+\log \left(2^2 \times 3\right) $
$=2 \log 3-\frac{1}{2} \times 4 \log 2+\log 2^2+\log 3 $
$ =2 \log 3-2 \log 2+2 \log 2+\log 3 $
$ =3 \log 3 $
$ =\log 3^3$
$ =\log 27 .$
View full question & answer→Question 124 Marks
Express the following as a single logarithm:$\log 144-\log 72+\log 150-\log 50$
Answer$\log 144 - \log 72 + \log 150 - \log 50$
$= \log (2^4 \times 3^2) - \log (2^3 \times 3^2) + \log (2 \times 3 \times 5^2) - \log (2 \times 5^2)$
$= \log 2^4+ \log 3^2 - {\log 2^3 + \log 3^2} + \log 2 + \log 3 + 5^2 - {\log 2 + \log 5^2}$
$= 4 \log 2 + 2 \log 3 - 3 \log 2 - 2 \log 3 + \log 2 + \log 3 + 2 \log 5 - \log 2 - 2 \log 5$
$= \log 2 + \log 3$
$= \log (2 x 3)$
$= \log 6.$
View full question & answer→Question 134 Marks
Write the logarithmic equation for:$V =\frac{1}{ D l} \sqrt{\frac{ T }{\pi r }}$
Answer$V =\frac{1}{ D l} \sqrt{\frac{ T }{\pi r }}$
$\Rightarrow V =\frac{1}{ D l}\left(\frac{ T }{\pi r }\right)^{\frac{1}{2}}$
Considering $\log$ on both the sides, we get
$\log V =\log \left[\frac{1}{ D l}\left(\frac{ T }{\pi r }\right)^{\frac{1}{2}}\right] $
$=\log \left(\frac{1}{ D l}\right)+\log \left(\frac{ T }{\pi r }\right)^{\frac{1}{2}} $
$=(\log 1-\log D -\log l)+\frac{1}{2} \log \left(\frac{ T }{\pi r }\right) $
$=(0-\log D -\log l)+\frac{1}{2}(\log T -\log \pi-\log r )$
$=\frac{1}{2}(\log T -\log \pi-\log r )-\log D -\log l .$
View full question & answer→Question 144 Marks
If $a = \log 20\ b = \log 25$ and $2\log (p - 4) = 2a - b,$ find the value of $'p\ '.$
Answer$a=\log 20, b=\log 25$ and $2 \log (p-4)=2 a-b$
$\Rightarrow 2 \log (p-4)=2 a-b$
$\Rightarrow 2 \log (p-4)=2 \log 20-\log 25$
$\Rightarrow \log (p-4)^2=\log 20^2-\log 25$
$\Rightarrow \log (p-4)^2=\log \left(\frac{400}{25}\right)$
$\Rightarrow(p-4)^2=\frac{400}{25}$
$\Rightarrow p^2-8 p+16=16$
$\Rightarrow p^2-8 p=0$
$\Rightarrow p(p-8)=0$
$\Rightarrow p=0$ or $p=8$
View full question & answer→Question 154 Marks
Write the logarithmic equation for:$n =\sqrt{\frac{ M \cdot g }{ m \cdot l }}$
Answer$n =\sqrt{\frac{ M \cdot g }{ m \cdot l}} $
$\Rightarrow n =\left(\frac{ M \cdot g }{ m \cdot l}\right)^{\frac{1}{2}}$
Considering $\log$ on both sides,
$\log n=\log \left(\frac{ M \cdot g }{ m \cdot l}\right)^{\frac{1}{2}} $
$=\frac{1}{2} \log \left(\frac{ M \cdot g }{ m \cdot l}\right) $
$=\frac{1}{2}[\log ( M \cdot g )-\log ( m \cdot l)] $
$=\frac{1}{2}(\log M +\log g -\log m -\log l) .$
View full question & answer→Question 164 Marks
If $a =\log \frac{ p ^2}{ qr }, b =\log \frac{ q ^2}{ rp }, c =\log \frac{ r ^2}{ pq }$, find the value of $a + b + c$.
Answer$a=\log \frac{p^2}{q r}, b=\log \frac{q^2}{r p}, c=\log \frac{r^2}{p q}$
Consider,
$ a+b+c$
$=\log \frac{p^2}{q r}+\log \frac{q^2}{r p}+\log \frac{r^2}{p q}$
$=\log p^2-\log q r+\log q^2-\log p+\log r^2-\log p q$
$=2 \log p-(\log q+\log r)+2 \log q-(\log r+\log p)+2 \log r-(\log p+\log q)$
$=2 \log p-\log q-\log r+2 \log q-\log r-\log p+2 \log r-\log p-\log q$
$=0 $
View full question & answer→Question 174 Marks
Prove that $\frac{\log _{ p } x}{\log _{ pq } x}=1+\log _{ p }$
Answer$ \text { L.H.S. }$
$=\frac{\log _{ p } x}{\log _{ pq } x}$
$=\frac{\left(\frac{\log x}{\log p }\right)}{\left(\frac{\log x}{\log q }\right)}$
$=\frac{\log x}{\log p } \times \frac{\log pq }{\log x}$
$=\frac{\log pq }{\log p }$
$=\frac{\log p +\log q }{\log p }$
$=1+\frac{\log q }{\log p }$
$=1+\log _{ p } q$
$=\text { R.H.S. } $
Hence proved.
View full question & answer→Question 184 Marks
Prove that $\log _{10} 125=3\left(1-\log _{10} 2\right)$
Answer$\text { L.H.S. }$
$=\log _{10} 125$
$=\log _{10}\left(\frac{1000}{8}\right)$
$=\log _{10} 1000-\log _{10} 8$
$=\log _{10}(10)^3-\log _{10}(2)^3$
$=3 \log _{10} 10-3 \log _{10} 2$
$=3 \times 1-3 \log _{10} 2$
$=3\left(1-\log _{10} 2\right)$
$=\text { R.H.S. }$
View full question & answer→Question 194 Marks
Express the following in terms of $\log 2$ and $\log 3: \log \frac{225}{16}-2 \log \frac{5}{9}+\log \left(\frac{2}{3}\right)^5$
Answer$\log \frac{225}{16}-2 \log \frac{5}{9}+\log \left(\frac{2}{3}\right)^5$
$=\log \frac{225}{16}-2 \log \frac{5}{9}+5 \log \frac{2}{3}$
$=\log 225-\log 16-2[\log 5-\log 9]+5[\log 2-\log 3]$
$=\log \left(5^2 \times 3^2\right)-\log 2^4-2\left[\log 5-\log 3^2\right]+5[\log 2-\log 3]$
$=\log 5^2+\log 3^2-4 \log 2-2[\log 5-2 \log 3]+5[\log 2-\log 3]$
$=2 \log 5+2 \log 3-4 \log 2-2 \log 5+4 \log 3+5 \log 2-5 \log 3$
$=\log 2+\log 3$
View full question & answer→Question 204 Marks
If $a=\log \frac{3}{5}, b=\log \frac{5}{4}$ and $c=2 \log \sqrt{\frac{3}{4}}$, prove that $5^{a+b-c}=1$
AnswerConsider $\log \left(5^{a+b-c}\right)$
$=(a+b-c) \log 5$
$=\left(\log \frac{3}{5}+\log \frac{5}{4}-2 \log \sqrt{\frac{3}{4}}\right) \log 5$
$=\left(\log \frac{3}{5}+\log \frac{5}{4}-\log \left[\sqrt{\frac{3}{4}}\right]^2\right) \log 5$
$=\left(\log \frac{3}{5}+\log \frac{5}{4}-\log \frac{3}{4}\right) \log 5$
$=\log \left(\frac{\frac{3}{5} \times \frac{5}{4}}{\frac{3}{4}}\right) \log 5$
$=\log 1 \times \log 5=0 \ldots[\because \log 1=0]$
$\therefore 5^{a+b-c}$
$=10^{\circ}$
$=1 .$
View full question & answer→Question 214 Marks
Express the following in terms of $\log 2$ and $\log 3: \log \sqrt[4]{648}$
Answer$\log \sqrt[4]{648}$
$=\log (648)^{\frac{1}{4}}$
$=\frac{1}{4} \log 648$
$=\frac{1}{4} \log \left(2^3 \times 3^4\right)$
$=\frac{1}{4} \log 2^3+\frac{1}{4} \log 3^4$
$=\frac{3}{4} \log 2+\frac{4}{4} \log 3$
$=\frac{3}{4} \log 2+\log 3$
View full question & answer→Question 224 Marks
Find the value of:$\frac{\log \sqrt{125}-\log \sqrt{27}-\log \sqrt{8}}{\log 6-\log 5}$
Answer$\frac{\log \sqrt{125}-\log \sqrt{27}-\log \sqrt{8}}{\log 6-\log 5}$
$=\frac{\log (125)^{\frac{1}{2}}-\log (27)^{\frac{1}{2}}-\log (8)^{\frac{1}{2}}}{\log 6-\log 5}$
$=\frac{\log (5)^{3 \times \frac{1}{2}}-\log (3)^{3 \times \frac{1}{2}}-\log (2)^{3 \times \frac{1}{2}}}{\log 6-\log 5}$
$=\frac{\frac{3}{2} \log (5)-\frac{3}{2} \log (3)-\frac{3}{2} \log (2)}{\log (2 \times 3)-\log 5}$
$=\frac{\frac{3}{2}[\log (5)-\log (3)-\log (2)]}{\log 2+\log 3-\log 5}$
$=\frac{\frac{3}{2}[\log (5)-\log (3)-\log (2)]}{-[\log 5-\log 3-\log 2]}$
$=-\frac{3}{2} .$
View full question & answer→Question 234 Marks
Find the value of:$\frac{\log \sqrt{8}}{8}$
Answer$\frac{\log \sqrt{8}}{8}$
$=\frac{\log 2 \sqrt{2}}{8}$
$=\frac{1}{8}(\log 2+\log \sqrt{2})$
$=\frac{1}{8}\left(\log 2+\log 2^{\frac{1}{2}}\right)$
$=\frac{1}{8} \log 2+\frac{1}{8} \log 2^{\frac{1}{2}}$
$=\frac{1}{8} \log 2+\frac{1}{2} \frac{1}{8} \log 2$
$=\frac{1}{8} \log 2+\frac{1}{16} \log 2$
$=\frac{2}{16} \log 2+\frac{1}{16} \log 2$
$=\frac{3}{16} \log 2 .$
View full question & answer→Question 244 Marks
If $x + \log 4 + 2 \log 5 + 3 \log 3 + 2 \log 2 = \log 108,$ find the value of $x.$
Answer$x+\log 4+2 \log 5+3 \log 3+2 \log 2=\log 108$
$\Rightarrow x =\log 108-\log 4-2 \log 5-3 \log 3-2 \log 2$
$=\log \left(2^2 \cdot 3^3\right)-\log 2^2-\log 5^2-\log 3^3-\log 2^2$
$=\log \left(\frac{2^2 \cdot 3^3}{2^2 \cdot 5^2 \cdot 3^3 \cdot 2^2}\right)$
$=\log \left(\frac{1}{100}\right)$
$\Rightarrow x$
$=\log 1-\log 100$
$=0-2$
$=-2$
$\therefore x=-2 \text {. }$
View full question & answer→Question 254 Marks
If $\log x^2-\log \sqrt{y}=1$, express $y$ in terms of $x$. Hence find $y$ when $x =2$.
Answer$\log x^2-\log \sqrt{y}=1$
$\Rightarrow \log \left(\frac{x^2}{\sqrt{y}}\right)=\log 10 $
$\Rightarrow \frac{x^2}{\sqrt{y}}=10 $
$\Rightarrow \sqrt{y}=\frac{x^2}{10}$
Squaring both sides, we get
$y=\left(\frac{x^2}{10}\right)^2=\frac{x^4}{100}$
Now, when $x =2$,
$y=\frac{2^4}{100}=\frac{16}{100}$
$=\frac{4}{25}$.
View full question & answer→Question 264 Marks
If $x^2+y^2=7 x y$, prove that $\log \left(\frac{x-y}{3}\right)=\frac{1}{2}(\log x+\log y)$
Answer$x ^2+ y ^2=7 xy$
$\Rightarrow x ^2+ y ^2-2 xy =7 xy -2 xy$
$\Rightarrow( x + y )^2=9 xy$
$\Rightarrow\left(\frac{x+y}{3}\right)^2= xy$
$\Rightarrow\left(\frac{x+y}{3}\right)=\sqrt{x y} $
Considering log both sides, we get
$ \log \left(\frac{x+y}{3}\right)=\log (x y)^{\frac{1}{2}}$
$\Rightarrow \log \left(\frac{x+y}{3}\right)=\frac{1}{2} \log (x y)$
$\Rightarrow \log \left(\frac{x+y}{3}\right)=\frac{1}{2}[\log x+\log y] . $
View full question & answer→Question 274 Marks
If $x^2+y^2=6 x y$, prove that $\log \left(\frac{x-y}{2}\right)=\frac{1}{2}(\log x+\log y)$
Answer$x^2+y^2=6 x y $
$\Rightarrow x^2+y^2-2 x y=6 x y-2 x y $
$\Rightarrow(x-y)^2=4 x y $
$\Rightarrow\left(\frac{x-y}{2}\right)^2=x y $
$\Rightarrow\left(\frac{x-y}{2}\right)=\sqrt{x y}$
Considering log both sides, we get
$\log \left(\frac{x-y}{2}\right)=\log (x y)^{\frac{1}{2}} $
$\Rightarrow \log \left(\frac{x-y}{2}\right)=\frac{1}{2} \log (x y) $
$\Rightarrow \log \left(\frac{x-y}{2}\right)=\frac{1}{2}[\log x+\log y] .$
View full question & answer→Question 284 Marks
If $\log 27 = 1.431,$ find the value of the following $:\log 300$
Answer$\log 27$
$=\log 3^3$
$=3 \log 3$
$=1.431$
$\Rightarrow \log 3$
$=\frac{1.431}{3}$
$=0.477$
$\therefore \log 300$
$=\log (3 \times 100)$
$=\log \left(3 \times 10^2\right)$
$=\log 3+2 \log 10$
$=0.477+2$
$=2.477 .$
View full question & answer→Question 294 Marks
If $\log 8 = 0.90$, find the value of each of the following: $\log4$
Answer$\log 4$
$\log 8$
$=\log 2^3$
$=3 \log 2$
$=0.90$
$\Rightarrow \log 2$
$=0.90 / 3$
$=0.3$
$\therefore \log 4$
$=\log 2^2$
$=2 \log 2$
$=2 \times 0.30$
$=0.60$
View full question & answer→Question 304 Marks
If $\log 2=0.3010, \log 3=0.4771$ and $\log 5=0.6990$, find the values of: $\log \sqrt{72}$
Answer$\log \sqrt{72} $
$=\log (72)^{\frac{1}{2}}$
$ =\frac{1}{2} \log 72 $
$ =\frac{1}{2} \log \left(2^3 \times 3^2\right) $
$=\frac{1}{2} \log 2^3+\frac{1}{2} \log 3^2 $
$=\frac{3}{2} \log 2+\frac{2}{2} \log 3$
$ =\frac{3}{2} \log 2+\log 3$
$ =\left(\frac{3}{2} \times 0.3010\right)+0.4771$
$ =0.9286 .$
View full question & answer→Question 314 Marks
If $2 \log x + 1 = 40$, find: $\log 5x$
Answer$2 \log x + 1 = 40$
$\Rightarrow 2\log x + \log10 = 40$
$\Rightarrow 2\log 10x = 40$
$\Rightarrow \log2 \times 5x = 20$
$\Rightarrow \log2 + \log5x = 20$
$\Rightarrow \log5x = 20 - \log2$
$\Rightarrow \log5x = 20 - 0.3010 ......($Since $\log2 = 0.3010)$
$\Rightarrow \log5x = 19.6989.$
View full question & answer→Question 324 Marks
If $\log _3 m=x$ and $\log _3 n=y$, write down $3^{1-2 y+3 x}$ in terms of $m$ an $n$
Answer$3^{1-2 y+3 x}$ in terms of $m$ an $n$
$\log _3 m=x$
$\Rightarrow m =3^{ x }$
$\log _3 n = y$
$\Rightarrow n =3^{ y }$
$\therefore 3^{1-2 y+3 x}$
$=3 \cdot 3^{-2 y} \cdot 3^{3 x }$
$=3 \cdot\left(3^y\right)^{-2} \cdot\left(3^x\right)^3$
$=3 n^{-2} \cdot m^3$
$=\frac{3 m^3}{n^2}$.
View full question & answer→Question 334 Marks
If $\log 16=a, \log 9=b$ and $\log 5=c$, evaluate the following in terms of $a, b, c: \log 2 \frac{1}{4}$
Answer$\log 16=a, \log 9=b$ and $\log 5=c $
$\log 4^2=a, \log 3^2=b$ and $\log 5=c$
$ 2 \log 4=a, 2 \log 3=b$ and $\log 5=c $
$ \log 4=\frac{a}{2}, \log 3=\frac{b}{2}$ and $\log 5=c $
Consider,$\log 2 \frac{1}{4}=\log \left(\frac{9}{4}\right) $
$ =\log 9-\log 4$
$ =\log 3^2-\log 4 $
$ =2 \log 3-\log 4 $
$ =2\left(\frac{b}{2}\right)-\frac{a}{2} $
$ =\frac{2 b-a}{2} .$
View full question & answer→Question 344 Marks
If $\log 16 = a, \log 9 = b$ and $\log 5 = c,$ evaluate the following in terms of $a, b, c: \log 720$
Answer$\log 16=a, \log 9=b$ and $\log 5=c$
$\log 4^2=a, \log 3^2=b$ and $\log 5=c$
$2 \log 4=a, 2 \log 3=b$ and $\log 5=c$
$\log 4=\frac{a}{2}, \log 3=\frac{b}{2}$ and $\log 5=c$
Consider, $\log 720=\log \left(4^2 \times 3^2 \times 5\right)$
$=\log 4^2+\log 3^2+\log 5$
$=2 \log 4+2 \log 3+\log 5$
$=2\left(\frac{a}{2}\right)+2\left(\frac{b}{2}\right)+c$
$=a+b+c$
View full question & answer→Question 354 Marks
If $\log 16 = a, \log 9 = b$ and $\log 5 = c,$ evaluate the following in terms of $a, b, c: \log 75$
Answer$\log 16=a, \log 9=b$ and $\log 5=c$
$\log 4^2=a, \log 3^2=b$ and $\log 5=c$
$2 \log 4=a, 2 \log 3=b$ and $\log 5=c$
$\log 4=\frac{a}{2}, \log 3=\frac{b}{2}$ and $\log 5=c$
Consider,$\log 75=\log \left(5^2 \times 3\right)$
$=\log 5^2+\log 3$
$=2 \log 5+\log 3$
$=2(c)+\frac{b}{2}$
$=\frac{4 c+b}{2} .$
View full question & answer→Question 364 Marks
If $\log 16 = a, \log 9 = b$ and $\log 5 = c,$ evaluate the following in terms of $a, b, c: \log 12$
Answer$\log 16=a, \log 9=b$ and $\log 5=c$
$\log 4^2=a, \log 3^2=b$ and $\log 5=c$
$2 \log 4=a, 2 \log 3=b$ and $\log 5=c$
$\log 4=\frac{a}{2}, \log 3=\frac{b}{2}$ and $\log 5=c$
Consider,$\log 12=\log (4 \times 3)$
$=\log 4+\log 3$
$=\frac{a}{2}+\frac{b}{2}$
$=\frac{a+b}{2} .$
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