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I solved graphically and found that $x + 3^x < 4$ is true for $x < 1$ but I can't find a way to prove it algebraiclly, any hints will be greatly appreciated!

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up vote 4 down vote accepted

The left hand side is an increasing function of $x$. With $x=1$, its value is $4$.

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I suppose we just take that $3^x$ is strictly increasing from the fact that $3^x = e^{x ln(3)}$ and the properties of $e$, right? – Edgar Sánchez Dec 9 '12 at 23:58
Yes, or with smaller teaspoons, that $3>1$ so that $\ln 3>0$, and the exponential function (i.e., base $e$) is strictly increasing. – Harald Hanche-Olsen Dec 10 '12 at 8:48

The solution is $x<1$

Since $x+3^x$ is strictly increasing. Therefore for all $x<1$ we have $x+3^x<1+3^1=4$.

If $1\leq x$ we have $1+3^1\leq x+3^x$ because $x+3^x$ is increasing.

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Thank you, that was fast! – Edgar Sánchez Dec 9 '12 at 23:58

Having found the solution $1$ to $x+3^x=4$ toucan use the fact that the derivative is positive to show it is unique.

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It's a pre-calculus problem so we are not allowed to use derivatives yet, but thank you anyway! – Edgar Sánchez Dec 9 '12 at 23:59

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