Need a hint for the problem $x + 3^x < 4$

Question

$x + 3^x < 4$

It's clear from looking at it that the answer is x < 1, but I want to solve it algebraically and that's what i'm having trouble with. I've messed around with it for a while and i've gotten nowhere. Logarithms just take me in a circle--for instance:

$3^x < 4 - x$

$\ln(3^x) < \ln(4-x)$

$x\ln(3) < \ln(4-x)$

$x<\frac{\ln(4-x)}{\ln(3)}$

$x < \log_3(4-x)$

$ 3^x < 4 - x$

and boom I'm back where I started.

I'm wondering if someone can just point me in the right direction. Thanks.

Answer 1

The best way is to look at the equation $x+3^x=4$, argue that $1$ is a solution and that since $x\to x+3^x$ is increasing it can only have one solution.

Answer 2

Note that $$f(x)=x+3^x\\f'=1+3^x.\ln3 >0$$is increasing function. so f(x)=4 has at least 1 root(solution) now you can write inequality in this form $$x+3^x<4\\x+3^3<1+3^1 \to x<1$$

Answer 3

I don't know what you mean exactly by ‘algebraically’, but you may observe the function is increasing (sum of two increasing functions – no derivative required here). As $1+3^1=4$, this means $x+3^x<4$ if $x<1$, $x+3^x>4$ if $x>1$.