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

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

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

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

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

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

$\log_n r= \frac{1}{log_r n}$,$\log_r n r\neq 1$,$\log_a1 =0$

Given $n^4<10^n$ - $\operatorname{far}$ fixed positive integer $n \geq 2$

Show that $(n+1)^4<10^{n+1}$.

Soln. $n^4<10^n \Longleftrightarrow \log _{10} n^4<\log _{10} 10^n$

$\Longleftrightarrow 4 \log _{10} n<n \log _{10} 10$

$4 \log _{10} n<n \longrightarrow$ (1)

$\frac{(n+1)^4}{10^{n+1}}<1 \quad(T \cdot P . T .)$

$\log _{10}\left[\frac{(n+1)^4}{10^{n+1}}\right]<\log _{10} 1=0$

$\log _{10}\left[\frac{(n+1)^4}{10^{n+1}}\right]<0$

$\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n$

$L H S =\log _{10}(n+1)^4-\log _{10} 10^{n+1}$

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

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

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

$\text { from (1)} < n+4 \log _{10}\left(1+\frac{1}{n}\right)- x-1$

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

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

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

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

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

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

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

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

$81<160 \Leftrightarrow \log _{10} 81<\log _{10} 160$

$\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}$

Show that $x^{y+z}+y^{z+x}+z^{x+y} \geqslant 3$

Soln. $\frac{\log _a x}{y-z}=\frac{\log _a y}{z-x}=\frac{\log _a z}{x-y}=k$

$\log _a x=k(y-z)$

$\log _a y=k(z-x)$

$\log _a z=k(x-y)$

$\log _a x=k(y-z) \Leftrightarrow x=a^{k(y-z)}$

$\log _a y=k(z-x) \Leftrightarrow y=a^{k(z-x)}$

$\log _a z=k(x-y) \Leftrightarrow z=a^{k(x-y)}$

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

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

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

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

$\log _{0.1}(\sin 2 x)+\log _{10}(\cos x)=\log _{10} 7$

Solve for $x$

Soln. $\log _{0.1}(\sin 2 x) =\log _{(10)^{-1}}(\sin 2 x)$

$\left.=-\log _{10}(\sin 2 x)\right)$

$\log _b m^a=\frac{1}{m} \log _b a$

$\log _{10}(\cos x)-\log _{10}(\sin 2 x)=\log _{10} 7$

$\log _{10}\left(\frac{\cos x}{\sin 2 x}\right)=\log _{10} 7$

$\frac{\cos x}{\sin 2 x}=7$

$\frac{\cos x}{2 \sin x \cos x}=7$

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

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

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

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

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

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

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

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

$x$ lies in the first quadrant

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

$\log _{\cos x} \sin x+\log _{\sin x} \cos x=2$

Solve for $x$ a > 0 a $\neq$ 1

Soln. $\frac{\log _a \sin x}{\log _a \cos x}+\frac{\log _a \cos x}{\log _a \sin x}=2$

$\left(\log _a \sin x\right)^2+\left(\log _a \cos x\right)^2-2\left(\log _a \cos x\right)\left(\log _a \sin x\right)=0$

$\left(\log _a \sin x-\log _a \cos x\right)^2=0$

$\log _a \sin x=\log _a \cos x$

$\Leftrightarrow \quad \sin x =\cos x$

$\Leftrightarrow \quad \tan x =1$

$x =\frac{\pi}{4}+n \pi$

$\sin x \neq 1 \quad \cos x \neq 1 \quad \sin x>0 \quad \cos x>0$

$x$ lies in the first quadrant

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

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