mathematics-class-12-unit-05-chapter-01-logarithm-

### <Success style="font-size:25px"> Logarithm L-1  </Success>
### <primary style="font-size:24px"> Introduction </primary>
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-  Every positive number b, can written in the exponential form as b = $a^x$ 
 
-  where $a>0, \underline{a} \neq 1$, then $\underline{a}$ is called as base, \& $x$ is called as exponent, power as index
 
-  Then the same expression can now be written as (known as logarithmic form) as
 
- $\log _a b=x$
 
- $\underline{a}^x=\underline{b} \quad \Leftrightarrow \log _a b=x \quad(a>0, a \neq 1, b>0)$
 
-  (i.e, $4=2^2, 9=3^2$, $27=3^3$ )
 
- $(\log _2  4=2, log _3 9=2, \log _3 27=3 .)$
 
- $ .\log _a b=x \Rightarrow \cdot a^x=b \quad(7 \text {-rule })$

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<footer style = "font-size: 18px"> $\rightarrow$ Logarithm lecture - 1 $\rightarrow$ <span style = "color:blue"> Introduction </span> $\rightarrow$ Problem 1 $\rightarrow$ Problem 2 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-1--success-3"><Success style="font-size:25px"> Logarithm L-1  </Success></h3>
<h3 id="primary-stylefont-size24px-problem-4-primary"><primary style="font-size:24px"> Problem 4 </primary></h3>
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<li>Find the value of $\rightarrow$ (a) $ \log_2$ 1 , (b) $ log_2$ 8 ,(c) $ \log_{81}$ 27 , (d) $ log_{\frac{1}{3}} 9\sqrt{3}$</li>
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<p>Solutions:</p>
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<p>(a) $ \log_2$ 1 = x (say), $ 2^x=1 \quad(7 \text {-rule }) $</p>
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<p>$\therefore 2^x=2^0 \quad x=0 \quad  \log_2 1 = x = 0 \quad a>0 \quad a^0 = 1$</p>
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<p>(b) $ \log_2$ 8 = x (say)  $ 2^x=8$ $\quad$ (can I write 8 as $2^x$?)$ \quad \therefore 2^x=2^3$ (By inspection)$\quad  \therefore x=3  $$ \log_2 8 = 3 $</p>
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<p>(c) $ \log_{81}$ 27 = x (say)$ 81^x =27$ $\quad \therefore(3^4)^x =3^3$ $\quad \therefore 3^{4 x} =3^3$ $ \quad \therefore 4x =3 \quad \therefore x = \frac{3}{4}$</p>
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<p>$27=3^3, 81=3^4 \quad$ (By inspection) $\left(a^m\right)^n=a^{m n}$$\log _{81} 27=\frac{3}{4}$</p>
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<p>(d) $ \log_{\frac {1}{3}} 9 \sqrt{3}=x \text {(say)}$     $(\frac{1}{3})^x =9 \sqrt{3}$ $\quad \therefore 3^{-x} =3^{5/2}$ $ \quad \therefore x = \frac {5}{2}$</p>
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<p>$ \therefore x =\frac{-5}{2}$,$\left(\frac{1}{a}\right)^m=a^{-m}$,$ 9 \sqrt{3} =3^2 \times 3^{1/2}$</p>
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<p>$ =3^{2+1/2}$,$ =3^{5/2}$,$a^m a^n=a^{m+n}~ \text {log}_{1/3}9\sqrt3 = \frac{-5}{2}$</p>
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<footer style = "font-size: 18px">Problem 2 $\rightarrow$ Problem 3 $\rightarrow$ <span style = "color:blue"> Problem 4 </span> $\rightarrow$ Observations $\rightarrow$ Problem 5 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-1--success-4"><Success style="font-size:25px"> Logarithm L-1  </Success></h3>
<h3 id="primary-stylefont-size24px-observations-primary"><primary style="font-size:24px"> Observations </primary></h3>
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<li>b$=1^x, \log _1$ b is not meaningful</li>
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<li>$\quad 1=1^x$ has infinitely many solutions. $\log _1 1$</li>
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<p>$\log _1$ b is not defined</p>
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<li>$b=a^{{\log _a b}} \quad a>0, a \neq 1, b>0$,$\log _a b=x($ say $)$</li>
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<p>$ a^x=b$,$ a^{\log _a b}=b$,$ 3^3=27 $,$ 3^{\log _3 27}=27$ (Verified)</p>
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<li>$\log _a b=x$,$\log _2 32=5$</li>
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<p>b is called ‘antilog’ of x to the base a</p>
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<p>$\text { antilog }_2 5 = 32$</p>
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<footer style = "font-size: 18px">Problem 3 $\rightarrow$ Problem 4 $\rightarrow$ <span style = "color:blue"> Observations </span> $\rightarrow$ Problem 5 $\rightarrow$ Solution </footer>

<Success style="font-size:25px; text-align:center;"> <b> Logarithm L-1 </b></Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p>${\log _3 \sqrt{\sqrt[3]{\sqrt[3]{3 \ldots}}}}$,$ {~ y = \sqrt{\sqrt[3]{\sqrt[3]{3 \ldots}}}}$</p>
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<p>$y=\sqrt{3 y}$,$~ y^2 = 3y$</p>
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<p>$\Leftrightarrow y^2-3 y=0$,$\Leftrightarrow y(y-3)=0$,$\Leftrightarrow y=0$ or $y=3$.$y=0$ not allowed</p>
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<p>$\log _3 y=\log _3 3 =1$,$\log _3 3=x(say)$,$\therefore 3^2=3^1$,$\therefore x = 1$</p>
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<p>$\log _a y \text { is defined only when } y>0$</p>
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<p>${\log _3 \sqrt{\sqrt[3]{\sqrt[3]{3 \ldots}}}} = \log_3 3 = 1$</p>
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<footer style = "font-size: 18px">Observations $\rightarrow$ Problem 5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Logarithm L-1  </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 3. $\log _{\sqrt{5}-2}(2+\sqrt{5})$
 
- $(\sqrt{5}-2)(\sqrt{5}+2)=5-4=1$

- $(\sqrt{5}+2)=\frac{1}{\sqrt{5}-2}$
 
- $\log _{\sqrt{5}-2}(2+\sqrt{5})$ $=\log _{(\sqrt{5}-2)}\left(\frac{1}{\sqrt{5}-2}\right)$
 
- $ = -1$
 
- $\log _a(1 / a)=x \Leftrightarrow a^x=\frac{1}{a}$
 
- $ \therefore a^x  =a^{-1}$

- $ \therefore x  =-1$
 
- $(a+b)(a-b)  =a^2-b^2$
 
- $\log _a(1 / a)=-1$
 
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-logarithm-l-1--success-6"><Success style="font-size:25px"> Logarithm L-1  </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
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<li>$8^{\log _8 x}+2 x+9=0$</li>
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<p>find the value of x, for which this expression is meaningful</p>
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<p>$a^{\log _a x}=x$</p>
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<p>$\text { Put } y=\log_ax$</p>
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<p>$a^y=x \Longleftrightarrow a^{\log _a} x=a^y=x$</p>
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<p>$ x+2 x+9=0$</p>
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<p>$ 3 x=-9 $</p>
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<p>$ x = \frac{-9}{3}=-3$</p>
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<p>$8^{\log _8(-3)}$ is not defined</p>
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<p>there is no solution for this expression</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Logarithm L-1  </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 5. $\log _{1/3}(a^2-1)=-1$
 
- Find the value of a : $(1 / 3)^{-1}=a^2-1$
 
- $ \therefore \quad 3^1  =a^2-1$,$ \therefore \quad 0  =(a^2-4)$

- $ \therefore \quad a  = \pm 2, ~ (a-2) (a+2 = 0) $,$ a  =2,-2$,$ a  =2$
 
- $(\frac{1}{a})^m=a^{-m}$
 
- Case 1.  $a=2 \quad \log _{\frac{1}{3}}(4-1)=-1$
 
- $\log _{1 / 3} 3=-1$
 
- Case 2. $a=-2 \quad \log _{1 / 3}(4-1)=-1$
 
- $\log _{1 / 3} 3=-1$.
 
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>