Chapter 7

$$ \sqrt{17-4 x}+\sqrt{x-5} \leq \sqrt{13 x+1} $$

$$ \log ^{2}(4-x)+\log (4-x) \cdot \log \left(x+\frac{1}{2}\right)-2 \log ^{2}\left(x+\frac{1}{2}\right)=0 $$

$$ \log _{x} 2-\log _{4} x+\frac{7}{6}=0 $$

$$ \log _{0.5 x} x^{2}-14 \log _{16 x} x^{3}+40 \log _{4 x} \sqrt{x}=0 $$

$$ 4 \log _{\frac{1}{2}} \sqrt{x}+2 \log _{4 x} x^{2}=3 \log _{2 x} x^{3} $$

$$ \log _{3 x}\left(\frac{3}{x}\right)+\log _{3}^{2} x=1 $$

$$ \sqrt{1+\log _{0.04} x}+\sqrt{3+\log _{0,2} x}=1 $$

$$ \sqrt{2-\log _{x} 9}=-\frac{\sqrt{12}}{\log _{3} x} $$

$$ \log _{x}\left(x^{2}+1\right)=\sqrt{\log _{\sqrt{x}}\left(x^{2}\left(1+x^{2}\right)\right)+4} $$

$$ \sqrt{\log _{2} x}-0.5=\log _{2} \sqrt{x} $$

$$ \log 2+\log \left(4^{x-2}+9\right)=1+\log \left(2^{x-2}+1\right) $$

$$ \log \left(6 \cdot 5^{x}+25 \cdot 20^{x}\right)=x+\log 5 $$

$$ \left|1-\log _{4} x\right|+1=\left|3-\log _{4} x\right| $$

$$ (x+1)^{\log (x+1)}=100(x+1) $$

$$ x^{\log x}=1000 x^{2} $$

$$ x^{\frac{\log x+5}{3}}=10^{5+\log x} $$

$$ |x-1|^{\log ^{2} x-\log x^{2}}=|x-1|^{3} $$

$$ |x-3|^{\frac{x+1}{4}}=|x-3|^{\frac{x-2}{3}} $$

$$ |x-3|^{3 x^{2}-10 x+3}=1 $$

$$ (\sqrt{x})^{\log _{5} x-1}=5 $$

$$ x^{\log x+7}=10^{4(\log x+1)} $$

$$ x^{3 \log x-\frac{1}{\log x}}=\sqrt[3]{10} $$

$$ 3 x^{2}-7 x+4 \leq 0 $$

$$ 3 x^{2}-7 x+6<0 $$

$$ 3 x^{2}-7 x-6<0 $$

$$ x^{2}-3 x+5>0 $$

$$ x^{2}-14 x-15>0 $$

$$ 2-x-x^{2} \geq 0 $$

$$ x^{2}-25<0 $$

$$ x^{2}+10 x \leq 7 x $$

$$ x^{2}-7 x<3 $$

$$ -x^{2}-16+8 x \geq 0 $$

$$ x^{2}+5 x+8>0 $$

$$ 5 x-1<(x+1)^{2}<7 x-3 $$

$$ x^{2}+2 \leq 3 x \leq 2 x^{2}-5 \text { . } $$

$$ x(x-1)^{2}>0 $$

$$ x(x-1)^{2} \leq 0 $$

$$ (2-x)(3 x+1)(2 x-3)>0 $$

$$ (3 x-2)(x-3)^{3}(x+1)^{3}(x+2)^{4}<0 $$

$$ x^{3}-64 x>0 $$

$$ x^{4}+8 x^{3}+12 x^{2} \geq 0 $$

$$ (x-1)\left(x^{2}-3 x+8\right)<0 $$

$$ (x-1)\left(x^{2}-1\right)\left(x^{3}-1\right)\left(x^{4}-1\right) \leq 0 $$

$$ \left(16-x^{2}\right)\left(x^{2}+4\right)\left(x^{2}+x+1\right)\left(x^{2}-x-3\right) \leq 0 $$

$$ \left(x^{2}-4\right)\left(x^{2}-4 x+4\right)\left(x^{2}-6 x+8\right)\left(x^{2}+4 x+4\right)<0 $$

$$ \left(2 x^{2}-x-5\right)\left(x^{2}-9\right)\left(x^{2}-3 x\right) \leq 0 $$

$$ x^{3}-3 x^{2}-10 x+24>0 $$

$$ x^{3}+4 x^{2}+5 x+2 \leq 0 $$

$$ 2 x^{3}-3 x^{2}+7 x-3>0 $$

$$ x^{4}-3 x^{3}+x^{2}+3 x-2 \geq 0 $$