Inequality with the variable in both the base and the exponent. [duplicate]

I realised that what I took as terseness of my question, actually made it look like a lazy attempt to get a homework answer. The following is the edited question, hopefully up to the standards of this great community.

I am in process of self-studying analysis using Spivak's calculus. The book is amazing, but the answers for problems are very brief, sometimes not sufficient for my limited knowledge. In this problem, I am interested in the mechanics of obtaining the reduced inequality (i.e. x

Find all values of x for which:

$\ 3^x+x < 4$

I managed to find the answer (x<1) using "brute-force", i.e. pondering what values of x would satisfy the equation, and obtaining the result without manipulating the formula, but my curiosity remains unsatisfied so as to the issue of how to manipulate this simple inequality to reduce it to solution.

marked as duplicate by Jonas Meyer, Chappers, Mark Bennet, mrf, Joel Reyes NocheJun 4 '15 at 0:25

• See this question. – Pim Jun 2 '15 at 5:30

As you can check, $$3^x+x<4$$ can be rewritten $$(4-x)\ln(3)e^{(4-x)\ln(3)}>81\ln(3),$$ or, setting $w=(4-x)\ln(3)$, $$we^w>81\ln(3).$$

Thus, using the main branch of the Lambert function, $$w>W(81\ln(3)),$$ and $$x<4-\frac{W(81\ln(3))}{\ln(3)}=1.$$

• Using the Lambert function is indeed overkill, but it will work for other constants. – Yves Daoust Jun 2 '15 at 16:02

Clearly $3^x+x=4$ when $x=1$. Moreover, if $x$ increases (decreases) then $3^x+x$ increases (decreases). So the solution of the inequality is $x<1$.

• Thanks! I should have specified it in the question, but I am looking for the algebraic manipulation that would get me there. The clever way to solve it works, but I am interested in the mechanics of it. – KubaWi Jun 2 '15 at 5:42
• Equations which mix exponential and polynomial terms are frequently very hard to solve. I would be surprised if there were any "systematic" solution to this problem. – David Jun 2 '15 at 5:44
• That's all I wanted to know. Thanks a lot, David! – KubaWi Jun 2 '15 at 5:53

You can show that $x < 1$. Indeed, if $x \geq 1 \Rightarrow 3^x+x \geq 3^1+1 = 4$, and if $x \leq 0 \Rightarrow 3^x + x \leq 3^0 + 0 = 1 < 4$. For if $0 < x < 1 \Rightarrow 3^x + x < 3^1 + 1 = 4$.

• Would you care to elaborate? – KubaWi Jun 2 '15 at 5:31