$\log_{10}({\dfrac{1}{70}})$= $\log_{10}{1}- \log_{10}{70}$
=>$- \log_{10}{7\times 10}$
=>-(a+1)
$\log_{10}{5}+\log_{10}{(5x+1)}$=$\log_{10}{(x+5)}+1$
=>$\log_{10}{5}+\log_{10}{(5x+1)}$=$\log_{10}{(x+5)}+\log_{10}{10}$
=>$\log_{10}{5(5x+1)}$
=>$\log_{10}{10(x+5)}$
=>5(5x+1)=10(x+5)
=>5x+1=2x+10
=>3x=9
=>x=3
Given expression:
$\log_{60}{3}+\log_{60}{4}+\log_{60}{5}$
=>$\log_{60}{(3\times 4\times 5)}$
=>$\log_{60}{60}$
=1.
Let $\log{2}{16}$=n.
then,$2^{n}$=16=$2^{4}$=>n=4
Therefore ,$\log{2}{16}$=4
$log \dfrac{a}{b}+log \dfrac{b}{a}$=log(a+b)
=>log(a+b)=$log(\dfrac{a}{b} \times \dfrac{b}{a})$=log 1.
so,a+b=1
$log_{2}[log_{3}(log_{2} x)]$ = 1
=>$log_{2}[log_{3}(log_{2} x)]$ = $log_{2}(2)$
=> $log_{3}(log_{2} x)$=2
=>$log_{2} x$=$3^2$=9
=> x = $2^9$ = 512
$\dfrac{1}{log_{xy}xyz}+\dfrac{1}{log_{yz}xyz}+\dfrac{1}{log_{zx}xyz}$
= $log_{xyz} xy + log_{xyz} yz + log_{xyz} zx$
=$log_{xyz}(xy \times yz \times zx)$
=$log_{xyz}(xyz)^{2}$
= $2log_{xyz}xyz$
= 2 x 1
= 2
$log_{a}b$ =$\dfrac{log\:b}{log\:a}, log_{b}c$ =$\dfrac{log\:c}{log\:b}, log_{c}a$ =$\dfrac{log\:a}{log\:c}$
$log_{a}b \times log_{b}c \times log_{c}a$ =$\dfrac{log\:b}{log\:a}\times\dfrac{log\:c}{log\:b}\times\dfrac{log\:a}{log\:c}$=$\dfrac{1}{2}\times\dfrac{1}{3}\times\dfrac{k}{5}$
=>1=$\dfrac{k}{30}$
=>k = 30
Given Exp. = log [{(9/8) / (27/32)} x 3/4)]
= log [(9/8) x (3/4) x (32/27)]
= log 1
= 0
$log_{5}\left(\dfrac{x^2+x}{x+1}\right)$=$log_{5}25$
=>$\left(\dfrac{x^2+x}{x+1}\right)$=25
=>$x^2-24x-25$=0
=>$(x-25)(x+1)$=0
=>x=25