We have a number $a > 1$ and we know the following inequality:
$$2^x \leq (ax)^4$$
And need to find an upper bound on $x$.
I thought of trying to calculate where $2^x$ intersects $(ax)^4$ and then the larger intersection would be an upper bound for $x$.
So this is what I did:
I called the value where they intersects $t$ and solved:
$$2^t = (at)^4\\
t\ln2 = 4\ln(at)\\
at\ln2=4a\ln(at)\\
\frac{\ln2}{4a}=(at)^{-1}\ln(at)\\
-\frac{\ln2}{4a}=(at)^{-1}\ln((at)^{-1})\\
-\frac{\ln2}{4a}=e^{\ln((at)^{-1})}\ln((at)^{-1})\\
W\left(-\frac{\ln2}{4a}\right)=\ln((at)^{-1})\\
t=\frac{e^{-W\left(-\frac{\ln2}{4a}\right)}}{a}$$
And therefore:
$$x\leq \max \left\{\frac{e^{-W_0\left(-\frac{\ln2}{4a}\right)}}{a},\frac{e^{-W_{-1}\left(-\frac{\ln2}{4a}\right)}}{a}\right\}$$
But I don't know how to continue from here. How can I bound this expression with $W$? I need a bound that doesn't use the W Lambert function, instead something like a non-infinite polinomyal or logarithmic function.
Best Answer
There are good bounds (have a look here) $$-1-\sqrt{2u}-u < W_{-1}(-e^{-(u+1)}) < -1-\sqrt{2u}-\frac{2}{3}u$$ For your case $$u=-\log \left(\frac{e }{4 a}\log (2)\right)$$