Logarithmic Inequalities

In the previous articles, we already learned about linear inequalities, and quadratic inequalities. But there are still more types of inequalities.

I’ll just discuss to you one, the logarithmic inequalities. The others will also be discussed as we go along the way.

First we recall logarithmic equations.

Let’s try to answer a problem.

Solve for x in:

\log (x+1)+\log (x-2)=1

Solution:

Using addition law for logarithm, we have

\log (x+1)+\log (x-2)=\log (x+1)(x-2)=\log (x^2-x-2)

We can rewrite our original equation as,

\log (x^2-x-2)=1

Recall that y=\log_bx \longleftrightarrow b^y=x

x^2-x-2=10^1

x^2-x-12=0

(x-4)(x+3)=0

x=4,-3

Now, substituting the values to the original equation, we can check that

x = -3 is an extraneous solution, and only x = 4 satisfies the equation. Therefore, the value of x is x = 4.

In solving logarithmic inequalities, we will also use that method. The rules will be shown when we will solve an example problem.

Problem:

Find the solution set in

\log (x-1)+\log (x-2)>1

Solution:

\log (x^2-x-2)>1

x^2-x-2>10

x^2-x-12>0

(x-4)(x+3)>0

Recalling quadratic inequalities,

x>4,x<-3 ♦

OOOPS!!! That’s not yet the final answer. 😀

Let’s go back to the equation. We have the terms log(x-1) and log(x-2). All real values of x cannot be the domain of the two terms. Remember again domain and range of logarithmic functions …

If \log A=x where x\epsilon \Re, therefore, A>0

So the domains of the two terms are:

\log (x-1)\to x-1>0\to x>1 ♦♦

\log (x-2)\to x-2>0\to x>2 ♦♦♦

Getting the intersection of ♦ , ♦♦, and ♦♦♦,

Our answer is

\boxed{x>4}

Worked Problem 1:

Solve the inequality   \log (x+1)-\log (x-1)\ge 2

[toggle title=”Solution:”]

\log\big (\displaystyle\frac{x+1}{x-1}\big )\ge 2

\displaystyle\frac{x+1}{x-1}\ge 100

x+1\ge 100x-100

101\ge 99x

99x\leq 101

x\leq \displaystyle\frac{101}{99}     @

But

x+1>0\to x>-1       @

x-1>0\to x>1      @@@

From @, @@, and @@@,

\boxed{1<x\leq\displaystyle\frac{101}{99}}

[/toggle]

 

Worked Problem 2:

Solve the inequality :   \log_2(x-3)^2+24>(\log_2(x-3))^2

[toggle title=”Solution:”]

\log_2(x-3)^2-24>(\log_2(x-3))^2

Let  A=\log_2(x-3). Recall that \log u^n\longleftrightarrow n\log u

2\log_2(x-3)+24>(\log_2(x-3))^2

2A+24>A^2

A^2-2A-24<0

(A-6)(A+4)<0

-4<A<6

But A=\log_2(x-3)

-4<\log_2(x-3)<6

\displaystyle\frac{1}{16}<x-3<64

\displaystyle\frac{49}{16}<x-3<67  @

But  x-3>0\to x>3  @@

From the intersection of @ and @@ we have,

\boxed{\displaystyle\frac{49}{16}<x<67}

[/toggle]

 

Worked Problem 3*:

If   x\epsilon [0,2\pi],  solve for x in:         \log_2(sinx)<-\displaystyle\frac{1}{2}

[toggle title=”Solution:”]

sinx<2^{-\frac{1}{2}}

sinx<\displaystyle\frac{\sqrt{2}}{2}

Looking at the graph of  f(x)=sinx

0\leq x<\displaystyle\frac{\pi}{4}, \displaystyle\frac{3\pi}{4}<x\leq 2\pi  @

But

sinx>0

0<x<\pi

Therefore,

0<x<\displaystyle\frac{\pi}{4}

[/toggle]

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