mathematics-class-12-unit-05-chapter-04-logarithm-jhd4velookm

<h3 id="success-stylefont-size25px-logarithm-l-4-success-1"><Success style="font-size:25px"> Logarithm L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-primary"><primary style="font-size:24px"> Problem 1 </primary></h3>
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<p>$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$</p>
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<p>To show $\quad x^{b+c} y^{c+a} z^{a+b}=1,~ $</p>
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<p>$\log_a1=0$</p>
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<p>Soln.   $\log_{10} (x^{b+c} y^{c+a} z^{a+b})=\log_{10}1, ~ $</p>
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<p>$ \log_x ( m n p ) $</p>
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<p>$\Leftrightarrow \log _{10}\left(x^{b+c}\right)+\log _{10}\left(y^{c+a}\right)+\log _{10}\left(z^{a+b}\right)=0 $</p>
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<p>$\Leftrightarrow(b+c) \log _{10} x+(c+a) \log _{10} y+(a+b) \log _{10} z=0 $</p>
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<p>$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=w $</p>
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<p>$\log _a m^n=n \log _a m$</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Logarithm lecture - 4 $\rightarrow$ <span style = "color:blue"> Problem 1 </span> $\rightarrow$ Solution $\rightarrow$  </footer>

<h3 id="success-stylefont-size25px-logarithm-l-4-success-3"><Success style="font-size:25px"> Logarithm L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-2-primary"><primary style="font-size:24px"> Problem 2 </primary></h3>
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<p>If $a^2=b^3=c^5=d^6$,</p>
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<p>show that $\log _d(a b c)=\frac{31}{5}$</p>
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<p>Soln.   $a^2=b^3=c^5=d^6=k$ (say)</p>
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<p>$a=k^{1 / 2}, b=k^{1 / 3}, c=k^{1 / 5}, d=k^{1 / 6}$</p>
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<p>$\log _d(a b c) =\log _{k^{1 / 6}}\left(k^{1 / 2} k^{1 / 3} k^{1 / 5}\right) $</p>
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<p>$=\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) $</p>
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<p>$=\log _{k^{1 / 6}}\left(k^{31 / 30}\right) $,$=\frac{31}{30} \log _{k^{1 / 6}} k$</p>
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<p>$=\frac{31}{30} \log _{k^{1 / 6} k} $,$=\frac{31}{30} \frac{\log _k k}{\log _k k^{1 / 6}} $</p>
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<p>$=\frac{31}{30} \frac{1}{(1 / 6) \log _k k} $,$=\frac{31}{30} \times \frac{6}{1} \times 1=\frac{31}{5}$</p>
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<p>(Change of base),$\left(\log _a b=\frac{\log _c b}{\log _c a}\right)$</p>
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<footer style = "font-size: 18px">Problem 1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 2 </span> $\rightarrow$ Problem 3 $\rightarrow$ Problem 4 </footer>

### <Success style="font-size:25px"> Logarithm L-4 </Success>
### <primary style="font-size:24px"> Problem 4 </primary>
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- $\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) $

- $=3 \times 1=3$

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<footer style = "font-size: 18px">Problem 2 $\rightarrow$ Problem 3 $\rightarrow$ <span style = "color:blue"> Problem 4 </span> $\rightarrow$ Problem 5 $\rightarrow$ Problem 6 </footer>

### <Success style="font-size:25px"> Logarithm L-4 </Success>
### <primary style="font-size:24px"> Problem 6 </primary>
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- $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)$

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<footer style = "font-size: 18px">Problem 4 $\rightarrow$ Problem 5 $\rightarrow$ <span style = "color:blue"> Problem 6 </span> $\rightarrow$ Problem 7 $\rightarrow$ Problem 8 </footer>

### <Success style="font-size:25px"> Logarithm L-4 </Success>
### <primary style="font-size:24px"> Problem 7 </primary>
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- $\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

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<footer style = "font-size: 18px">Problem 5 $\rightarrow$ Problem 6 $\rightarrow$ <span style = "color:blue"> Problem 7 </span> $\rightarrow$ Problem 8 $\rightarrow$ Problem 9 </footer>

### <Success style="font-size:25px"> Logarithm L-4 </Success>
### <primary style="font-size:24px"> Problem 8 </primary>
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- $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

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

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

- $=\log _{a b c}(a b c)=1$

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<footer style = "font-size: 18px">Problem 6 $\rightarrow$ Problem 7 $\rightarrow$ <span style = "color:blue"> Problem 8 </span> $\rightarrow$ Problem 9 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Logarithm L-4 </Success>
### <primary style="font-size:24px"> Problem 9 </primary>
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- $\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

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<footer style = "font-size: 18px">Problem 7 $\rightarrow$ Problem 8 $\rightarrow$ <span style = "color:blue"> Problem 9 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>