Question

Решите неравенство (2^x)*(3^1/x)>6

Answer 1

$\Big2^\big{x}*\Bigg3^{\dfrac{1}{x}}>6\\\\ODZ\::\;\;\;\;\boxed{\;\;x\neq0\;\;}\\\\log_2\bigg(\Big2^\big{x}*\Bigg3^{\dfrac{1}{x}}\bigg)>log_26$

$Formyla\;:\\\\\boxed{\;\;log_a\Big(bc\Big)=log_ab+log_ac\;\;}\\\\\\\boxed{\;\;log_ab^n=n*log_ab\;\;}\\\\log_2\Big(\Big2^\big{x}\Big)+log_2\Big(\bigg3^{\dfrac{1}{x}}\Big)>log_23+1\\\\x+\dfrac{1}{x}*log_23>log_23+1\\\\\big(x-1\big)+log_23*\Big(\;\dfrac{1}{x}-1\;\Big)>0\\\\\big(x-1\big)+log_23*\Big(\;\dfrac{1-x}{x}\;\Big)>0\\\\\big(x-1\big)*\bigg(1-\dfrac{log_23}{x}\bigg)>0\\\\\big(x-1\big)*\bigg(\dfrac{x-log_23}{x}\bigg)>0$

Метод интервалов :

---------------(0)++++++++++(1)-----------(log₂3)++++++++> x

$Otvet\;:\;\;\boxed{\;\;\Big(0\;;1\Big)\;\;\cup\;\;\Big(log_23;+\infty\Big)\;\;}$