# Logarithmic inequality

• Jun 7th 2010, 09:27 AM
rowe
Logarithmic inequality
Solve $\displaystyle x+3^x < 4$ for x.

Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x-4<0$, giving me what I suspect is right, x < 1. But how to be sure?
• Jun 7th 2010, 10:30 AM
earboth
Quote:

Originally Posted by rowe
Solve $\displaystyle x+3^x < 4$ for x.

Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x-4<0$, giving me what I suspect is right, x < 1. But how to be sure?

1. Graph the function $\displaystyle f(x)=x+3^x-4$ and check for which values of x the values of f(x) are negative.

2. attempt: Solve
$\displaystyle x+3^x=4$ for x. Since $\displaystyle y = x \text{ and } y = 3^x$ are monotically increasing there exists only one solution. By first inspection x = 1.
Therefore the inequality is true for all x < 1.
• Jun 7th 2010, 10:51 AM
Quote:

Originally Posted by rowe
Solve $\displaystyle x+3^x < 4$ for x.

Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x-4<0$, giving me what I suspect is right, x < 1. But how to be sure?

Alternatively,

$\displaystyle x+3^x\ <\ 4$

$\displaystyle 3^x\ <\ 4-x$

$\displaystyle f(x)=4-x$ is a line decreasing at $\displaystyle 45^o,$ passing through
the y-axis at (0,4) and the x-axis at (4,0).

$\displaystyle g(x)=3^x$ is an increasing function passing through the y axis at (0,1), so it cuts f(x) between x=0 and x=4.

When x=1, both f(x) and g(x) are 3.

$\displaystyle g(x)\ <\ f(x)\ for\ x\ <\ 1$
• Jun 7th 2010, 10:53 AM
p0oint
Quote:

Originally Posted by earboth
1. Graph the function $\displaystyle f(x)=x+3^x-4$ and check for which values of x the values of f(x) are negative.

2. attempt: Solve
$\displaystyle x+3^x=4$ for x. Since $\displaystyle y = x \text{ and } y = 3^x$ are monotically increasing there exists only one solution. By first inspection x = 1.
Therefore the inequality is true for all x < 1.

@earboth

1. it would be much easier if he draw 3^x and 4-x and find values where 4-x is "bigger" then 3^x

2. I do not what made you think that if they're "monotically increasing there exists only one solution"?

example: x^3+5 and 3^x, both monotonically increasing but have 2 intersections.
• Jun 7th 2010, 10:59 AM
Soroban
Hello, rowe!

Quote:

Solve: .$\displaystyle x+3^x \:<\: 4$
There is no algebraic solution, but there is a graphical solution.

We have: . $\displaystyle 3^x \:<\:4-x$

The question becomes: when is $\displaystyle y = 3^x$ below $\displaystyle y = 4-x$ ?

The graphs look like this:
Code:

|
* |    *
*
| *  *
|  o
| * : *
*  :  *
*  |  :    *
*        |  :      *
- - - - - - + - + - - - - * - -
|  1          *
|

As you pointed out, the curves intersect at $\displaystyle x = 1.$

You answer is correct: . $\displaystyle x \:<\:1$