Question

Find the solution set in ${\log }_2(x-1)+{\log }_2(x-2)>1$

• A. x>1
• B. x>2
• C. x>3
• D. ($-\infty$,0) U (3,$\infty$)

Answer 1

The correct option is C

x>3

${\log }_2(x-1)+{\log }_2(x-2)>1$

For Logarithm to be defined

x - 1 > 0, x > 1 ------------------(1)

And x - 2 > 0, x > 2 ------------(2)

This inequality can also be written as

${\log }_2(x-1)(x-2)$ > 1

${\log }_2(x^2-3x+2)$ > 1

Since, base of the logarithm greater than 1. So, given logarithm is a decreasing function.

$x^2-3x+2>2^1$

$x^2-3x+2-2>0$

$x^2-3x>0$

$\left(x\right)(x-3)>0$ $x\in (-\infty ,0)\cup (3,\infty )$ ---(3)

So,

From equation 1,2 and 3 Or common part of equation 1,2 and 3

We get,

x > 3