Question

1) log за основою одна третя (2х+5)менше 0 2) log за основою одна четверта(3-4х)більше -1 3) log за основою 16(4х+3)більше одної другої 4) lg (12-5х) менше 0

Answer 1

1.

$\\\log_{\frac{1}{3}}(2x+5)<0\\ 2x+50\\ 2x-5\\ x-\frac{5}{2}\\\\ \log_{\frac{1}{3}}(2x+5)<\log_{\frac{1}{3}}\left(\frac{1}{3}\right)^0\\ 2x+5{\left(\frac{1}{3}\right)}^0\\ 2x+51\\ 2x-4\\ x-2\\\\ x-\frac{5}{2} \wedge x-2\\ \underline{x-2}$

2.

$\\\log_{\frac{1}{4}}(3-4x)-1\\ 3-4x0\\ -4x-3\\ x<\frac{3}{4}\\\\ \log_{\frac{1}{4}}(3-4x)\log_{\frac{1}{4}}\left(\frac{1}{4}\right)^{-1}\\ 3-4x<{\left(\frac{1}{4}\right)}^{-1}\\ 3-4x<4\\ -4x<1\\ x-\frac{1}{4}\\\\ x<\frac{3}{4} \wedge x-\frac{1}{4}\\ \underline{x\in(-\frac{1}{4},\frac{3}{4})}$

3.

$\\\log_{16}(4x+3)\frac{1}{2}\\ 4x+30\\ 4x-3\\ x-\frac{3}{4}\\\\ \log_{16}(4x+3)\log_{16}16^{\frac{1}{2}}\\ 4x+316^{\frac{1}{2}}\\ 4x+34\\ 4x1\\ x\frac{1}{4}\\\\ x\frac{1}{4} \wedge x-\frac{3}{4}\\ \underline{x\frac{1}{4}}$

4.

$\\\lg(12-5x)<0\\ 12-5x0\\ -5x-12\\ x<\frac{12}{5}\\\\ \lg(12-5x)<\lg10^0\\ 12-5x<10^0\\ 12-5x<1\\ -5x<-11\\ x\frac{11}{5}\\\\ x\frac{11}{5} \wedge x<\frac{12}{5}\\ \underline{x\in(\frac{11}{5},\frac{12}{5})}$