Question

помогите ,прошу вас,последняя надежда на вас((((

Answer 1

$4.\;log_{3sqrt[4]3}left(27cdotsqrt[2]3right)+log_{4sqrt[2]{32}}left(8cdotsqrt[2]2right)=\ =log_{3cdot3^{frac14}}left(3^3cdot3^{frac12}right)+log_{2^2cdot2^{frac52}}left(2^3cdot2^{frac12}right)=\ =log_{3^{frac54}}left(3^{frac72}right)+log_{2^frac92}left(2^{frac72}right)=frac45log_3left(3^{frac72}right)+frac29log_2left(2^{frac72}right)=\ =frac45cdotfrac72+frac29cdotfrac72=frac{14}5+frac79=frac{126+35}{45}=frac{161}{45}=3frac{26}{45}$

$5.\1);log_6(x+1,5)=-log_6x\ log_6(x+1,5)+log_6x=0\ log_6(x+1,5)x=0\ x^2+1,5x=1\ x^2+1,5x-1=0;times2\ 2x^2+3x-2=0\ D=9+4cdot4cdot2=25\ x_1=-2,;x_2=0,5\ 2);log_3log_8log_2(x-5)=log_32-1\ log_3log_8log_2(x-5)=log_32-log_33\ log_3log_8log_2(x-5)=log_3frac23\ log_8log_2(x-5)=frac23\ log_2(x-5)=8^{frac23}=sqrt[3]{8^2}=sqrt[3]{64}=4\ x-5=2^4\ x-5=16\ x=21$

$3);log_3(x-5)=log_32+frac12log_3(3x-20)\ log_3(x-5)=log_32+log_3(3x-20)^{frac12}\ log_3(x-5)=log_3left(2sqrt{3x-20}right)\ x-5=2sqrt{3x-20}\ x^2-10x+25=4(3x-20)\ x^2-10x+25=12x-80\ x^2-22x+105=0\ D=484-4cdot105=64\ x_1=7,;x_2=15\ 4);log_5(log(sqrt{x^2+19}))=0\ log(sqrt{x^2+19})=5^0\ log_{10}(sqrt{x^2+19})=1\ sqrt{x^2+19}=10\ x^2+19=100\ x^2=81\ x_1=-9,;x_2=9$