What values of x satisfy the inequality x + 3^x < 4?

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The inequality x + 3^x < 4 is satisfied for values of x less than 1. The function f(x) = x + 3^x is strictly increasing, meaning that once it reaches 4 at x = 1, it will exceed 4 for any x greater than 1. As x approaches negative infinity, f(x) approximates y = x, while for larger x, it approaches y = 3^x. Therefore, the solution to the inequality is x < 1. This analysis highlights the behavior of the function and confirms the range of x that satisfies the inequality.
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Homework Statement



For what x does x + 3^x &lt; 4


Homework Equations





The Attempt at a Solution



The thing is = 4 when x is 1. So I want x < 1.
But is there a way to do this algebraically? Like with log or something?
 
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Taking a log doesn't really make this problem much more transparent. If this were the Calculus forum, I'd suggest looking at Newton's method for finding a numerical approximation: http://en.wikipedia.org/wiki/Newton's_method
 
You can say something about this inequality just by knowing something about the graphs of y = x and y = 3x. Both of these are strictly increasing functions, so their sum is also a strictly increasing function. Let's define f(x) = x + 3x.

Looking at asymptotic behavior, as x gets more and more negative, f(x) approaches the graph of y = x. IOW, for very negative x, f(x) \approx x. As x gets larger and larger, f(x) approaches y = 3x.

You already found out that f(1) = 4, so for any x > 1, then f(x) > 4. Similarly, if x < 1, f(x) < 4.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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