Logarithmic Inequalities

Question

Find the values of x for which 2ln(2x-3) > ln(x-1).

Answer 1

The first step to solving this inequality is finding its domain.

ln(x) is only defined for (0, +∞) or x>0

2ln(2x-3) is defined when (2x-3)>0, simplifying we get x>3/2

ln(x-1) is defined when (x-1)>0, simplifying we get x>1

To solve the given inequality both the functions must exist, so we must find the values of x for which both the logarithmic functions are defined.

When we merge the overlapping intervals of x>1 and x>3/2 we get x>3/2. You can use a number line to find the overlapping interval.

So our first condition is x>3/2 which we will call eq(1).

Now we can solve for the inequality, 2ln(2x-3) > ln(x-1)

Firstly, we will rewrite 2ln(2x-3) as ln(2x-3)², this step is to get equivalent expressions on both sides.

ln(2x-3)² > ln(x-1) both sides are a log function with the same base e.

Plotting the graph of ln(x), we can notice that as x increases ln(x) also increases.

Therefore (2x-3)² > (x-1)

4x²-12x+9 > x-1, now adding both sides with (1-x) we get

4x²-13x+10 > 0

Factorizing we get (4x-5)(x-2) > 0

So x < 5/4 or x>2 which we will call eq(2)

Finally we will find the overlapping intervals of eq(1) and eq(2). Plotting these intervals on a number line we get the interval x > 2. Hence 2ln(2x-3) > ln(x-1) for x > 2.