Solve $\displaystyle x+3^x < 4$ for x.
Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x4<0$, giving me what I suspect is right, x < 1. But how to be sure?
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Solve $\displaystyle x+3^x < 4$ for x.
Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x4<0$, giving me what I suspect is right, x < 1. But how to be sure?
1. Graph the function $\displaystyle f(x)=x+3^x4$ 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.
Alternatively,
$\displaystyle x+3^x\ <\ 4$
$\displaystyle 3^x\ <\ 4x$
$\displaystyle f(x)=4x$ is a line decreasing at $\displaystyle 45^o,$ passing through
the yaxis at (0,4) and the xaxis 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.
$\displaystyle g(x)\ <\ f(x)\ for\ x\ <\ 1$
@earboth
1. it would be much easier if he draw 3^x and 4x and find values where 4x 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.
Hello, rowe!
There is no algebraic solution, but there is a graphical solution.Quote:
Solve: .$\displaystyle x+3^x \:<\: 4$
We have: . $\displaystyle 3^x \:<\:4x$
The question becomes: when is $\displaystyle y = 3^x$ below $\displaystyle y = 4x$ ?
The graphs look like this:Code:
*  *
*
 * *
 o
 * : *
* : *
*  : *
*  : *
      +  +     *  
 1 *

As you pointed out, the curves intersect at $\displaystyle x = 1.$
You answer is correct: . $\displaystyle x \:<\:1$