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Find all $x$ for which $x+3^x<4$

I'm stuck at this one...how does one solve for $x$?

I've tried:

$x+3^x<4$

$3^x<4-x$

$x<\log_3({4-x})$

But I don't know where to go from there.

If I start by subtracting $3^x$ from each side:

$x+3^x<4$

$x<4-3^x$

I still don't know how to handle the $3^x$ term when there is another $x$ term. Help?

Edward Jiang
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Juanma Eloy
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  • It would improve your Question to state the domain from which possible $x$ will be taken. – hardmath Jan 23 '15 at 02:30
  • It didn't really specify that in the problem. I took it from the prologue of Spivak's calculus book. I suppose the domain should be all real numbers? I know how to operate with complex numbers too, but I'm not sure if there are any complex solutions to this problem. – Juanma Eloy Jan 23 '15 at 02:52
  • If we are using an inequality, real numbers seems more likely than complex numbers. However one might get different answers if restricting to integers (or natural numbers), etc. – hardmath Jan 23 '15 at 03:01

4 Answers4

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Let $f(x) = x + 3^x$, then $f'(x) > 0$ for all $x$. Also $f(1) = 4$; however for any $x < 1$,

$$f(x) = x + 3^x < 1 + 3 = 4$$

Hence ...

Simon S
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Use that $0+3^0=2<4 ; 1+3^1=4 \not>4$ and that $\frac {d}{dx} (x+3^x)=\frac{d}{dx}(x+e^{xln3})=1+(ln3)3^x>0 \forall x$.

user192680
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If $x=1$, then it would be $1+3^1<4$, which equals $1+3<4=4<4$, which isn't true, so $x<1$. The lower the exponent $x$ for base $3$ in the inequality, the less the left side is, so that's why.

ReliableMathBoy
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$x$ is an increasing function on $\mathbb{R}$. $3^x$ is an increasing function on $\mathbb{R}$. It follows that $x+3^x$ is an increasing function on $\mathbb{R}$.

At $x=1$, $x+3^x=4$. It follows by the definition of an increasing function, that for all $x<1$, $x+3^x<4$, and for all $x>1$, $x+3^x>4$.

2'5 9'2
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