mathematics-class-12-unit-05-chapter-05-logarithm-xgi056dnylo

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Problem 1 </primary>
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- Prove that $\log _{34} 5$ lies in $\left(\frac{1}{3}, \frac{1}{2}\right)$.

- $\text { ie. to prove } \frac{1}{3}<log _{34} 5<\frac{1}{2}$

- Soln.  $\frac{1}{3}<log _{34} 5$ & $log _{34} 5<1 / 2$,$1<{3 log _{34} 5}$ & ${2 log _{34} 5}<1$

- $n\log _a m=\log _a\left(m^n\right)$,$\text { i.e. } 1<log _{34} 5^3$ & $log _{34} 5^2<1$

- i.e. 1<$log_{34}{5}^3$ & $log_{34} {5}^2<1$,$log_a{a}=1$ a>0, a # 1

- $log_{34}{34}<log_{34}{125}$ & $log_{34}{25}<log_{34}{34}$,

- $m<n \Longleftrightarrow \log _a m<\log _a n$

- 34<125 $\Longleftrightarrow \log_{34}34 < \log_{34}125$

- 25<34 $\Longleftrightarrow \log_{34}25 < \log_{34}34$,$\frac{1}{3} < \log_{34}5 < \frac{1}{2}$

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<footer style = "font-size: 18px"> $\rightarrow$ Logarithm lecture - 5 $\rightarrow$ <span style = "color:blue"> Problem 1 </span> $\rightarrow$ Problem 2 $\rightarrow$ Problem 3 </footer>

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Problem 2 </primary>
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- Show that $\sum_{r=2}^{45} \frac{1}{\log _{r} n}=\log _n(45) !$

- m! = m(m-1)....2x1.

- Soln.  (Change of base),$log_b a =\frac{log_c a}{log_c b}$,$c>0, $ $c \neq 1$

- $\log_n(45)! = log_n(45)(44)....(2)(1)$

- Multiplication law

- $\log_a(mn)=log_am + log_a n$,$\log_a(mn) = \log_a m + \log_a n$

- $\log_n(45)! = \log_n 45 + \log_n 44 +.... + \log_n^2 + \log_n 1$,$=\sum_{r=1}^{45} \log_n r$

- $=\sum_{r=2}^{45} \log_nr$

- $=\sum_{r=2}^{45} \frac{1}{log_r n}$

- $\log_n r= \frac{1}{log_r n}$,$\log_r n r\neq 1$,$\log_a1 =0 $
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<footer style = "font-size: 18px">Logarithm lecture - 5 $\rightarrow$ Problem 1 $\rightarrow$ <span style = "color:blue"> Problem 2 </span> $\rightarrow$ Problem 3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ L H S  =\log _{10}(n+1)^4-\log _{10} 10^{n+1}$

- $=4 \log _{10}(n+1)-(n+1)\left(\log _{10} 10\right)$

- $=4 \log _{10}\left(n\left(1+\frac{1}{n}\right)\right)-(n+1)$

- $=4 \log _{10} n+4 \log _{10}\left(1+\frac{1}{n}\right)-n-1$

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

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\text { from (1)} < n+4 \log _{10}\left(1+\frac{1}{n}\right)- x-1 $

- $ n \geqslant 2 \Leftrightarrow \frac{1}{2} \geqslant \frac{1}{n} \Leftrightarrow\left(1+\frac{1}{2}\right) \geqslant\left(1+\frac{1}{n}\right)$

- $\Leftrightarrow \log _{10}\left(1+\frac{1}{2}\right) \geqslant \log _{10}\left(1+\frac{1}{n}\right)$

- $\text { LHS }<4 \log _{10}\left(1+\frac{1}{n}\right)-1 $,$\leqslant 4 \log _{10}\left(1+\frac{1}{2}\right)-1$

- $\text { LHS } <4 \log _{10}\left(1+\frac{1}{n}\right)-1$,$\leqslant 4 \log _{10}\left(1+\frac{1}{2}\right)-1$

- $=4 \log _{10}\left(\frac{3}{2}\right)-1$,$=\log _{10}\left(\frac{3}{2}\right)^4-\log _{10} 10$

- $=\log _{10}\left[\frac{81}{16 \times 10}\right]$

- $ =\log _{10}\left(\frac{81}{160}\right)$

- $=\log _{10} 81-\log _{10} 160<0$

- $81<160 \Leftrightarrow \log _{10} 81<\log _{10} 160$
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<footer style = "font-size: 18px">Problem 3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Sulution </span> $\rightarrow$ Problem 4 $\rightarrow$ Soluton </footer>

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\frac{1}{3}\left(a^{k(y-z)(y+z)}+a^{k(z-x)(z+x)}+a^{k(x-y)(x+y)}\right)$

- $=\frac{1}{3}\left(a^{k\left(y^2-z^2\right)}+a^{k\left(z^2-x^2\right)}+a^{k\left(x^2-y^2\right)}\right)$

- A.M. $\geqslant$ G.M.
- $\geqslant\left(a^{k\left(y^2-z^2\right)} a^{k\left(z^2-x^2\right)} a^{\left.k\left(x^2-y^2\right)\right)^{1 / 3}}\right.$

- A.M. $\geqslant$ G.M.
- $\geqslant\left(a^{k\left(y^2-z^2\right)} a^{k\left(z^2-x^2\right)} a^{\left.k\left(x^2-y^2\right)\right)^{1 / 3}}\right)$

- $=\left(a^{k\left[y^2-z^2+z^k-x^2+x^2-y^k\right]}\right)^{1 / 3}$

- = 1
- $\frac{1}{3}\left(x^{y+z}+y^{z+x}+z^{x+y}\right) \geq 1$

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

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \frac{1}{\sin x}=14 $

- $ \sin x=\frac{1}{14} $

- $ x=\sin ^{-1}\left(\frac{1}{14}\right) $

- $ x=n \pi+(-1)^n \sin ^{-1}\left(\frac{1}{14}\right) $

- $\sin 2 x>0$ & $\cos x>0$

- $ x=n \pi+\left((-1)^n \sin ^{-1}\left(\frac{1}{14}\right)\right) $

- $\sin 2 x>0 $  &  $ \cos x>0 $

- $\sin x>0 $  & $\cos x>0 $

- $x$ lies in the first quadrant

- $x=2 m \pi+\sin ^{-1}\left(\frac{1}{14}\right)$

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

### <Success style="font-size:25px"> Logarithm L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\sin x \neq 1 \quad \cos x \neq 1 \quad \sin x>0 \quad \cos x>0$

- $x$ lies in the first quadrant

- $x \neq n \pi+(-1)^n \frac{\pi}{2}, 2 n \pi$

- $x=2 m \pi+\frac{\pi}{4}$

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