Janet Forbes

2022-07-08

What is the number of real roots of $(\mathrm{log}x{)}^{2}-\lfloor \mathrm{log}x\rfloor -2=0$ $\lfloor \phantom{\rule{thinmathspace}{0ex}}\cdot \phantom{\rule{thinmathspace}{0ex}}\rfloor $represents the greatest integer function less than or equal to x.

I know how to solve logarithm equation but due to greatest integer function I am unable to proceed further please help thanks.

I know how to solve logarithm equation but due to greatest integer function I am unable to proceed further please help thanks.

Wade Atkinson

Beginner2022-07-09Added 12 answers

Since $[\mathrm{log}x]\le \mathrm{log}x$

we have $(\mathrm{log}x{)}^{2}-\mathrm{log}x-2\le 0$

This is equivalent to $-1\le \mathrm{log}x\le 2$

When $-1\le \mathrm{log}x\le 0,[\mathrm{log}x]=-1$ so that $\mathrm{log}x=\pm 1$ If we see that $\mathrm{log}x=1$ is not in the specified range. Hence $\mathrm{log}x=-1$ and $x=\frac{1}{10}$

When $0\le \mathrm{log}x<1$, $[\mathrm{log}x]=0$ so that $\mathrm{log}x=\pm \sqrt{2}$ None of these values in the range.

Similarly we can use $1\le \mathrm{log}x<2$ this will give us $x={10}^{\sqrt{3}}$

When $\mathrm{log}x=2$, $[\mathrm{log}x]=2$ and equation is satisfied. Thus $x=100$ is third real root.

we have $(\mathrm{log}x{)}^{2}-\mathrm{log}x-2\le 0$

This is equivalent to $-1\le \mathrm{log}x\le 2$

When $-1\le \mathrm{log}x\le 0,[\mathrm{log}x]=-1$ so that $\mathrm{log}x=\pm 1$ If we see that $\mathrm{log}x=1$ is not in the specified range. Hence $\mathrm{log}x=-1$ and $x=\frac{1}{10}$

When $0\le \mathrm{log}x<1$, $[\mathrm{log}x]=0$ so that $\mathrm{log}x=\pm \sqrt{2}$ None of these values in the range.

Similarly we can use $1\le \mathrm{log}x<2$ this will give us $x={10}^{\sqrt{3}}$

When $\mathrm{log}x=2$, $[\mathrm{log}x]=2$ and equation is satisfied. Thus $x=100$ is third real root.

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