Question

Помогите пожалуйста решить неравенство с логарифмами ​

Answer 1

Ответ:

$5^{lgx}-3^{lgx}<5\dfrac{1}{3}\cdot 3^{\frac{1}{2}\, lgx}\cdot 5^{\frac{1}{2}\, (lgx-2)}\ \,\ \ \ ODZ:\ x>0\ ,\\\\\\5^{lgx}-3^{lgx}<\dfrac{16}{3}\cdot 3^{lg\sqrt{x}}\cdot 5^{\frac{1}{2}\, lgx-1}\\\\\\\star \ \ lgx=lg(\sqrt{x})^2=2\, lg\sqrt{x}\ \ \star \\\\\\5^{2\, lg\sqrt{x}}-3^{2\, lg\sqrt{x}}<\dfrac{16}{3}\cdot 3^{lg\sqrt{x}}\cdot 5^{lg\sqrt{x}-1}\\\\\\\Big(5^{lg\sqrt{x}}\Big)^2-\Big(3^{lg\sqrt{x}}\Big)^2<\dfrac{16}{3}\cdot 3^{lg\sqrt{x}}\cdot 5^{lg\sqrt{x}}\cdot \dfrac{1}{5}$

$\Big(5^{lg\sqrt{x}}\Big)^2-\Big(3^{lg\sqrt{x}}\Big)^2<\dfrac{16}{15}\cdot 3^{lg\sqrt{x}}\cdot 5^{lg\sqrt{x}}\ \Big|:(3^{lg\sqrt{x}})^2>0\\\\\\\left(\Big(\dfrac{5}{3}\Big)^{lg\sqrt{x}}\right)^2-1<\dfrac{16}{15}\cdot \Big(\dfrac{5}{3}\Big)^{lg\sqrt{x}}\\\\\\t=\Big(\dfrac{5}{3}\Big)^{lg\sqrt{x}}>0\ \ \ \Rightarrow \ \ \ t^2-1<\dfrac{16}{15}\cdot t\ \ \ ,\ \ \ 15t^2-16t-15<0\ \ ,\\\\\\D/4=289=17^2\ ,\ \ t_1=-\dfrac{3}{5}<0\ \ ,\ \ t_2=\dfrac{5}{3}>0$

$znaki:\ \ +++(-\dfrac{3}{5})---(\dfrac{5}{3})+++\ \ \ ,\ \ \ -\dfrac{3}{5}<t<\dfrac{5}{3}\ \ ,\\\\t>0\ \ \ \Rightarrow \ \ \ 0<t<\dfrac{5}{3}\\\\\\0<\Big(\dfrac{5}{3}\Big)^{lg\sqrt{x}}<\dfrac{5}{3}\ \ \ \Rightarrow\ \ \ lg\sqrt{x}<1\ \ ,\ \ \sqrt{x}<10^1\ \ ,\ \ x<100\ \ ,\\\\\\{}\ \ \ \ \underline{0<x<100}$