Lg(3*5^x+24*20^x)=x+lg18 45 баллов

$\displaystyle lg(3*5^x+24*20^x)=x+lg18$

$\displaystyle lg(3*5^x+24*5^x*4^x)=lg 10^x+lg18$

$\displaystyle lg(3*5^x+24*5^x*4^x)=lg(18*10^x)$

$\displaystyle 3*5^x+24*5^x*4^x=18*10^x 3*5^x(1+8*4^x-6*2^x)=0$

$\displaystyle 3*5^x \neq 0$

$\displaystyle 1+8*2^{2x}-6*2^x=0$

$\displaystyle 2^x=t$

$\displaystyle 8t^2-6t+1=0 D=36-32=4=2^2 t_1=1/4 t_2=1/2$

$\displaystyle 2^x= \frac{1}{4} x=-2$

$\displaystyle 2^x= \frac{1}{2} x=-1$

Ответ х=-1 или х=-2