Question

Найдите корень уравнения

1)  (1/81)^х-8=2

2) log6(х+4)= log6(4х-2)

3)(1/3)^x-13=1/9

 

Answer 1

$(frac{1}{81})^{x-8}=2$

$3^{-4(x-8)}=2$

$log_33^{-4(x-8)}=log_32$

$-4(x-8) log_33=log_32$

$-4(x-8)=log_32$

$x-8=-frac{1}{4}log_32$

$x=-frac{1}{4}log_32+8$

 

$log_6(x+4)= log_6(4x-2)$

ОДЗ $left { {{x+4>0} atop {4x-2>0}} right. <=> left { {{x>-4} atop {4x>2}} right. <=> left { {{x>-4} atop {x>0,5}} right.$

ОДЗ x>0.5

x+4=4x-2

3x=6

x=2

 

$(frac{1}{3})^{x-13}=frac{1}{9}$

$(frac{1}{3})^{x-13}=(frac{1}{3})^2$

x-13=2

x=15