Question

1)log16x+log8x+log4x=13/2 2)log3 10 *log7 3 *logx7 =1 3))logx18-logx0.5 =2

Answer 1

1) $log _{16} x+log _{8}x+log _{4}x= \frac{13}{2}$
$log _{2 ^{4} } x+log _{2 ^{3} }x+log _{2 ^{2} }x= \frac{13}{2}$
$\frac{1}{4} log _{2}x+ \frac{1}{3} log _{2}x+ \frac{1}{2}log _{2} x= \frac{13}{2}$
$\frac{13}{12}log _{2} x= \frac{13}{2}$
$log _{2} x=6$
$x =2 ^{6} = 64$
2) $log _{3}10 *log _{7}3*log _{x} 7=1$
$\frac{log _{3}10 }{log _{3}7 } *log _{x} 7=1$
$log _{7} 10*log _{x} 7= 1$
$log _{7} 10= \frac{1}{log _{x} 7}$
$log _{7}10=log _{7}x$
$x=10$
3) $log _{x} 18-log _{x}0,5=2$
$log _{x} \frac{18}{0,5} = 2$
$log _{x} 36=2$
$x^{2} =36$
$x = 6$            $x=-6$ - посторонний корень ,так как -6 < 0