• $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$

• To show $\quad x^{b+c} y^{c+a} z^{a+b}=1,~$

• $\log_a1=0$

• Soln. $\log_{10} (x^{b+c} y^{c+a} z^{a+b})=\log_{10}1, ~$

• $\log_x ( m n p )$

• $\Leftrightarrow \log _{10}\left(x^{b+c}\right)+\log _{10}\left(y^{c+a}\right)+\log _{10}\left(z^{a+b}\right)=0$

• $\Leftrightarrow(b+c) \log _{10} x+(c+a) \log _{10} y+(a+b) \log _{10} z=0$

• $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=w$

• $\log _a m^n=n \log _a m$

• $\frac{\log _{10} x}{b-c}=\frac{\log _{10} y}{c-a}=\frac{\log _{10} z}{a-b}= w$

• $\log _{10} x=w(b-c)$

• $\log _{10} y=w(c-a)$

• $\log _{10} z=w(a-b)$

• Putting $\circledast \circledast$ in $\circledast$

• $(b+c)(b-c) w+(c+a)(c-a) w+(a+b)(a-b) w=0$

• $w\left(b^2-c^2+c^2-a^2+a^2-b^2\right)=0$

• $w\cdot 0=0$

• If $a^2=b^3=c^5=d^6$,

• show that $\log _d(a b c)=\frac{31}{5}$

• Soln. $a^2=b^3=c^5=d^6=k$ (say)

• $a=k^{1 / 2}, b=k^{1 / 3}, c=k^{1 / 5}, d=k^{1 / 6}$

• $\log _d(a b c) =\log _{k^{1 / 6}}\left(k^{1 / 2} k^{1 / 3} k^{1 / 5}\right)$

• $=\log _{k^{1 / 6}}\left(k^{1 / 2+1 / 3+1 / 5}\right)$,$=\log _{k^{1 / 6}}\left(k^\frac{15+10+6}{30}\right)$

• $=\log _{k^{1 / 6}}\left(k^{31 / 30}\right)$,$=\frac{31}{30} \log _{k^{1 / 6}} k$

• $=\frac{31}{30} \log _{k^{1 / 6} k}$,$=\frac{31}{30} \frac{\log _k k}{\log _k k^{1 / 6}}$

• $=\frac{31}{30} \frac{1}{(1 / 6) \log _k k}$,$=\frac{31}{30} \times \frac{6}{1} \times 1=\frac{31}{5}$

• (Change of base),$\left(\log _a b=\frac{\log _c b}{\log _c a}\right)$

• $\frac{\log _2 5}{\log _2 11}-\frac{\log _4 5}{\log _4 11}$

• Soln. $= \log _{11} 5-\log _{11} 5=0$ .

• $\log _v \sqrt[4]{u^3} \cdot \log _w v^5 \cdot \log _u \sqrt[5]{w^4}=3 ?$

• Soln. $\text { LHS }=\log _v u^{3 / 4} \cdot \log _w v^5 \cdot \log _u w^{4 / 5}$

• $=\left(\frac{3}{4} \times 5 \times \frac{4}{5}\right)\left(\log _v u \cdot \log _w v \cdot \log _u w\right)$

• $=3\left(\log _v u \cdot \log _w v \cdot \log _u w\right)$

• $=3\left(\frac{\log _a u}{\log _av} \cdot \frac{\log _a v}{\log _a w} \frac{\log _a w}{\log _a u}\right)$

• $=3 \times 1=3$

$\log _6 7 =\frac{\log _2 7}{1+\log _2 3}$

• Soln. $LHS =\log _6 7$

• $=\frac{\log _2 7}{\log _2 6}$

• $=\frac{\log _2 7}{\log _2(2 \times 3)}$

• $=\frac{\log _2 7}{\log _2 2+\log _2 3}$

• $=\frac{\log _2 7}{1+\log _2 3}$

• $x=\log _{12} 6 \quad y=\log _{18} 12$ & $z=\log _{24} 18$

• To show $\quad 1+x y z=2 y z$

• Soln. $LHS =1+x y z$

• $=1+\log _{12} 6 \cdot \log _{18} 12 \cdot \log 18$

• $=1+\frac{\log 6}{\log 12} \cdot \frac{\log 12}{\log 18} \cdot \frac{\log 18}{\log 24}$

• $=1+\frac{\log 6}{\log 24}$,$=1+\log _{24} 6$

• $=\log _{24} 24+\log _{24} 6 \quad(1=\log _{24} 24)$

• $=\log _{24}(24 \times 6)=\log _{24}(144)$,

• RHS $=2 y z$

• $=2 \log _{18} 12 \cdot \log _{24} 18$

• $=2 \frac{\log 12}{\log 18} \frac{\log 18}{\log 24}$,$=2 \log _{24} 12=\log _{24}(144)$

• $\frac{1}{\log _{a b} {a b c}}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{{a c}} a {b c}}=2 \text { (P.T.) }$

• $\times\left(\log _a b=\frac{\log _b b}{\log _b a}=\frac{1}{\log _b a}\right)$

• Soln. $LHS =\log _{a b c} a b+\log _{a b c} b c+\log _{a b c} a c$

• $=\log _{a b c}(a b \cdot b c \cdot a c)$

• $=\log _{a b c}\left(a^2 b^2 c^2\right)$

• $=\log _{a b c}(a b c)^2$

• $=2 \log _{a b c}(a b c)=2=$ RHS

• $x=1+\log _a(b c)=\log _a a+\log _a(b c)=\log _a(a b c)$

• $y=1+\log _b(a c)=\log _b b+\log _b(a c)=\log _b(a b c)$

• $z=1+\log _c(a b)=\log _c c+\log _c(a b)=\log _c(a b c)$

• show that $x y+y z+z x=x y z$

• Soln. $\frac{x y+y z+z x}{x y z} =1$

• $\therefore \quad \frac{1}{z}+\frac{1}{x}+\frac{1}{y} =1$

• Substituting the values of $x, y$ & $z$, we get

• $\frac{1}{\log _c(a b c)}+\frac{1}{\log _a(a b c)}+\frac{1}{\log _b(a b c)}=1$

• $=\log _{a b c} c+\log _{a b c} a+\log _{a b c} b$

• $=\log _{a b c}(a b c)=1$

• $\log _a x+\log _{a^2} x^2+\log _{a^3} x^3 +\cdots+\log _{a^p} x^p=p \log _a x .$

• Soln. $\log _{a^m} b^n=n \log _{a^m} b$

• $=n \frac{1}{\log _b a^m}$

• $=\frac{n}{m} \frac{1}{\log _b a}$

• $=\frac{n}{m} \log _a b .$

• $\log _{a^m} b^n=\frac{n}{m} \log _a b$

• $\log _{a^m} x^m=\frac{m}{m} \log _a x=\log _a x$

• $\log _a x+\frac{2}{2} \log _a x+\frac{3}{3} \log _a x+\ldots+\frac{p}{p} \log _a x$

• $=(1+1+\ldots+1) \log _a x$

• $=p \log _a x=$ RHS