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 $\left \{ {{x+40} \atop {4x-20}} \right. <= \left \{ {{x-4} \atop {4x2}} \right. <= \left \{ {{x-4} \atop {x0,5}} \right.$

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