Exponential Inequalities

It’s me again, Lesley Arthur Dudley. J

We are already done with Linear, Quadratic, Polynomial, and Logarithmic inequalities. For those who didn’t read those articles yet, you can find them in this blog.

Now we are in our next stop, Exponential Inequalities.

Before that, we must review our Exponential Equations by answering some problems.

1). Solve for x such that 3^x=243

3^x=243

3^x=3^5

x=5

2). Solve for x such that 4^x-5\cdot 2^x+6=0.

Let 2^x=A,   so  4^x=(2^x)^2=A^2

A^2-5A+6=0

(A-3)(A-2)=0

A=2,3

2^x=2,3

x=1,\log_23

3). The half-life of a radioactive substance is 60 years. How much part will be left from the original substance after the first 20 years?

The function of the amount left is  S=Pe^{kt}
Where S is the amount remaining, P be the initial amount, e is the Euler’s constant, t be the time (in years) and k be a constant.

Expressing the half life,

\displaystyle\frac{P}{2}=Pe^{60k}\to e^{60k}=\displaystyle\frac{1}{2}

Unknown=Pe^{20k}

Notice that  (e^{60k})^{\frac{1}{3}}=e^{20k}

Thus,     e^{20k}=\big(\displaystyle\frac{1}{2}\big)^{\frac{1}{3}}=\displaystyle\frac{1}{\root 3\of{2}}=\displaystyle\frac{\root 3\of{2}}{4}

Therefore, \displaystyle\frac{\root 3\of{2}}{4} of the original substance will be left after 20 years.

Now let’s move on to Exponential Inequalities.

Here are the steps in solving exponential inequalities:

  • Transform the inequality into an equation first to solve for the variable unknown in the problem
  • Proceed as that in solving linear, quadratic, or polynomial inequalities to solve for the inequality

Sample problems:

1). Solve for x: 2^x<\displaystyle\frac{1}{4}

2^x<\displaystyle\frac{1}{4}

2^x=2^{-2}

x<-2

2). 4^x-2^x>2

2^x=4,4^x=A^2

A^2-A>2

A^2-A-2>0

But since A>0,

A>2\to 2^x>2

2^x>2^1\to x>1

 3). For what values of x is 4^x(6^x-36)(2^x+3)<0

Since 4^x and 2^x+3>0 for all real numbers x,

6^x-36<0\to <6^2

x<2

 4). 9^x+6^x<4^x

9^x+6^x<4^x\to 1+\big(\displaystyle\frac{2}{3}\big)<\big(\displaystyle\frac{4}{9}\big)^x

A=\big(\displaystyle\frac{2}{3}\big)^x

1+A<A^2

A^2-A-1<0

\big(A-\displaystyle\frac{1-\sqrt{5}}{2}\big)\big(A-\displaystyle\frac{1-\sqrt{5}}{2}\big)<0

But since  A>0 and \displaystyle\frac{1-\sqrt{5}}{2}<0A-\displaystyle\frac{1-\sqrt{5}}{2}>0  for all A and x.

So,   0<A<\displaystyle\frac{1+\sqrt{5}}{2}\to x<\log_{\frac{2}{3}}\big(\displaystyle\frac{1+\sqrt{5}}{2}\big)

This is just a short discussion of the topic. You may answer more math problems like these problems in this discussion by visiting my page in facebook, Lesley Arthur Dudley. And of course, don’t forget to like.

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