10 вычислить log2(4x-3) + log1/8(125) = log0.5(x) + log4(0.04)

Question

10 вычислить log2(4x-3) + log1/8(125) = log0.5(x) + log4(0.04)

Answer 1

0; x>0;\\\\ x>\frac{3}{4};\\\\ log_2 (4x-3)+log_{2^{-3}} 5^3=log_{2^{-1}} x+log_2^2 0.2^2;\\\\ log_2 (4x-3)-log_{2} 5=-log_{2} x+log_2 0.2;\\\\ log_2 \frac{4x-3}{5}=log_2 \frac{0.2}{x};\\\\ \frac{4x-3}{5}=\frac{0.2}{x};\\\\ (4x-3)x=0.2*5;\\\\ 4x^2-3x=1;\\\\ 4x^2-3x-1=0;\\\\ 4x^2-4x+x-1=0;\\\\ 4x(x-1)+1*(x-1)=0;\\\\ (4x+1)*(x-1)=0;\\\\ 4x+1=0;x_1=-\frac{1}{4}<\frac{3}{4};\\\\ x-1=0;x_2=1;x=1" alt="log_2 (4x-3)+log_{\frac{1}{8}} 125=log_{0.5} x+log_4 0.04;\\\\ 4x-3>0; x>0;\\\\ x>\frac{3}{4};\\\\ log_2 (4x-3)+log_{2^{-3}} 5^3=log_{2^{-1}} x+log_2^2 0.2^2;\\\\ log_2 (4x-3)-log_{2} 5=-log_{2} x+log_2 0.2;\\\\ log_2 \frac{4x-3}{5}=log_2 \frac{0.2}{x};\\\\ \frac{4x-3}{5}=\frac{0.2}{x};\\\\ (4x-3)x=0.2*5;\\\\ 4x^2-3x=1;\\\\ 4x^2-3x-1=0;\\\\ 4x^2-4x+x-1=0;\\\\ 4x(x-1)+1*(x-1)=0;\\\\ (4x+1)*(x-1)=0;\\\\ 4x+1=0;x_1=-\frac{1}{4}<\frac{3}{4};\\\\ x-1=0;x_2=1;x=1" align="absmiddle" class="latex-formula">

Answer 2

ОДЗ:4x-3>0           4x>3      x>3/4

         x>0                  x>0        x>0 

log2(4x-3)+log2 в -3 степени(125)=log2 в -1 степени(x)+log2 во 2степени(0,04)

log2(4x-3)-1/3log2(5 в 3 степени)=-log2(x)+1/2log2(0.2 во 2 степени)

log2(4x-3)-log2(5)=log2(0.2)-log2(x)

log2((4x-3)/5)=log2(0.2/x)

(4x-3)/5=0.2/x

(4x-3)x=0.2*5

4xx-3x=1

4xx-3x-1=0

D=9+4*4*1=25

x1=(3+5)/8=1

x2=(3-5)/8=-1/4 Не подходит по ОДЗ

 

Ответ;1