Question

Решите задание по алгебре 10-11 класс)

Answer 1

$a)100^2^x^+^1 =1 \ \ 100^2^x^+^1=100^0 \ \ 2x+1=0 \ \ 2x=-1 \ \ x=1/2 = 0,5 \ \ b)9cdot :81^{1-2x}=27^{2-x} \ \ 9cdot :81^{1-2x}=left(3^3right)^{2-x} \ \ 9left(3^4right)^{1-2x}=left(3^3right)^{2-x} \ \ 9cdot :3^{4left(1-2xright)}=3^{3left(2-xright)} \ \ 3^2cdot :3^{4left(1-2xright)}=3^{3left(2-xright)} \ \ 3^{2+left(4left(1-2xright)right)}=3^{3left(2-xright)} \ \ 2+left(4left(1-2xright)right)=3left(2-xright)$

$2+4left(1-2xright)=3left(2-xright) \ -8x+6=6-3x \ 8x=-3x \ -8x+3x=-3x+3x \ -5x=0 \ x=0$

$B)2^{2x}-14-2^x-32=0 \ \ (2^x)^2-14-2^x-32=0 \ t^2-14-t-32=0 \ left(-1right)^2-4cdot :1left(-46right)=185 \ \ t= frac{1б sqrt{185} }{2}$
$2^x=frac{1+ sqrt{185}}{2} ;x=log_2(1+ sqrt{185})-1$

2^x= 1 - sqrt(185) / 2  , x∈∅

$OTB:x=log_2(1+ sqrt{185})-1$

$G)2^{4x+3}-3cdot :2^{4x-1}-5cdot :2^{4x+1}=-56 \ \ (2^4-3-5*2^2)*2^4^x^-^1=-56 \ \ (16-3-5*4)*2^4^x^-^1=-56 \ \ (16-3-20)*2^4^x^-^1=-56 \ \ -7*2^4^x^-^1=-56 \ 2^4^x^-^1=8 \ \ 2^4^x^-^1=2^3 \ \ 4x-1=3 \ 4x=3+1 \ 4x=4 \ x=1$