Question

Найди "лишнее" слово: 1) слонёнок 3) медвежонок 2) детёныш 4) волчонок

Answer 1

$1. \ a) \ log_{ \frac{1}{ \sqrt{3} } }9^{3 \sqrt{3} }=log_{3^{-1/2}}3^{2*3 \sqrt{3} }= \frac{1}{-1/2} log_{3}3^{6 \sqrt{3} }=-2*6 \sqrt{3}log_33= \\ =-12 \sqrt{3} \\ b) \ 7^{2log_72+1}=7*7^{2log_72}=7*7^{log_72^2}=7*7^{log_74}=7*4=28 \\ 2. \ a) \ log_5(4x+1)=2-log_5(2x+3) \\ ODZ: \left \{ {{4x+10} \atop {2x+30}} \right. = \left \{ {{x-1/4} \atop {x-3/2}} \right.=x-1/4 \\ log_5(4x+1)+log_5(2x+3)=2 \\ log_5(4x+1)(2x+3)=2 \\ 8 x^{2} +12x+2x+3=25 \\ 8 x^{2} +14x-22=0$
$4 x^{2} +7x-11=0 \\ D=47+4*4*11=225=15^2 \\ x_{1,2}= \frac{-7 \pm 15}{8}=1;-11/4 \\ = x=1 \\ b) \ lg^2x-3lg(10x)=1 \\ ODZ:x0 \\ lg^2x-3(lg10+lgx)=1 \\ lgx=t \\ t^2-3(1+t)=1 \\ t^2-3t-4=0 \\ t_1=-4;t_2=1 \\ lgx=1=x_1=10 \\ lgx=-4=x_2=10^{-4}=0,0001$

$3. \ log_5x \leq 27-x \\ ODZ:x0\\ x=25=log_5{25}\leq 27-25=2 \leq 2 \\ \\ x \in (0;25]$

$4. \ x^{log_6x^2}+6^{log_6^2x}=42 \\ ODZ:x0 \\ x^{2*log_6x}+6^{log_6x*log_6x}=42 \\(x^{log_6x})^2+(6^{log_6x})^{log_6x}}=42 \\ ( \frac{1}{x^{log_x6}} )^2+6^{log_6x}}=42 \\ x+ (\frac{1}{6})^2 =42 \\ x=42- \frac{1}{36} \\ x= \frac{1511}{36}$