Question

solve the inequality of log x square base 2 X + 3 less than 1​

Answer 1

Answer:

value of x is less than 1 /√2

Step-by-step explanation:

Given:

$log_{2x}\,x^2+3<1$

$\frac{log\,x^2}{log\,2x}+3<1$

$\frac{2log\,x}{log\,2x}+3<1$

$\frac{2log\,x}{log\,2+log\,x}+3<1$

$\frac{2log\,x+3log\,2+3log\,x}{log\,2+log\,x}<1$

$5log\,x+3log\,2<log\,2+log\,x$

$5log\,x-1log\,x<log\,2-3log\,2$

$4log\,x<-2log\,2$

$log\,x<\frac{-2log\,2}{4}$

$log\,x<\frac{-log\,2}{2}$

$log\,x<log\,2^{\frac{-1}{2}}$

$x<2^{\frac{-1}{2}}$

$x<\frac{1}{\sqrt{2}}$

Therefore, value of x is less than 1 /√2