Solve in $\mathbb{R}^+$ algebra calculus Equation with derivatives

Question

I was trying to solve this question:

Solve in $\mathbb{R}^+$ the Equation

$7^x-5^x=\lfloor x^2 \rfloor+1$

Where $\lfloor t\rfloor$ is the floor function

It is known that $\lfloor x^2\rfloor\leq x^2 $ so

$$7^x-5^x<x^2+1 $$

but You hace

$$7^x-5^x-x^2>1$$

for $x>1 $

But it is hard with derivatives

Answer 1

You are on the right track. Let call $$ f(x)=7^x-5^x-x^2 $$ we have $$ f'(x)=\log(7) 7^x-\log(5) 5^x -2x $$ and $$ f^{''}(x)=(\log(7))^2 7^x-(\log(5))^2 5^x -2 $$

$f''(x)$ is increasing and $f''(1)>0$ so for $x \ge 1$ $f'(x)$ is increasing.

$f'(1)>0$, so for $x \ge 1$ $f(x)$ is increasing, but $f(1)=1$ that implies that, for any $x >1$ $$ 7^x-5^x-x^2>1 $$ This show that there are no solution for $x>1$. $x=1$ is a solution.

So it only remains to find the solution for $0\le x<1$ i.e. the solution of the equation $$ g(x):=7^x-5^x-1=0 $$ As $g(0)=-1$ and $g(1)=1$ and the function is continuous, by Bolzano theorem there is at least one solution.

As $g(x)$ is increasing this solution is unique

Edit:

If you are not allowed to use a calculator you can prove that $ f'(1)>0$ in the following way: We have $$ \log(7)7-\log(5) 5 > \log(7) (7-5)>2 $$ as $\log(7)> \log(5)>1$. So $f'(1)>0$ Similarly $$ (\log(7))^27-(\log(5))^2 5 > (\log(7))^2 (7-5)>2 $$
and so $f''(1)>0$

Answer 2

Take apart intervals $$x<0 \to 7^x-5^x<0 \ \text{ but } \ [x^2]+1\ge0 \ \text{ no solution}$$now for positives $$[x^2]=0,1,2,3,4,...\to x=0,1,\sqrt2 ,\sqrt3 , \sqrt4,...\\ 0\le x<1 \to \ [x^2]=0 \to 7^x-5^x=0+1 \ \text{ one solution}\\ 1\le x<\sqrt2 \to 7^x-5^x=1+1 \to x=1 \ \text{ one solution}\\ \sqrt2 \le x<\sqrt3 \to 7^x-5^x=2+1 \ \text{ no solution}\\ \sqrt3 \le x<\sqrt4 \to 7^x-5^x=3+1 \ \text{ no solution} \\ \vdots$$ by approximating $7^{\sqrt2}-5^{\sqrt2}\sim 7^{1.4}-5^{1.4}>5.3\ne2+1 $
by approximating $7^{\sqrt3}-5^{\sqrt3}\sim 7^{1.7}-5^{1.7}>12\ne3+1 $
and so on (there is no more solution)

Answer 3

It is evident that $x=1$ is solution.By convexity of functions $f(x)=7^x-5^x$ and $g(x)=x^2+1$ this intersection is unique. However, because of $f(x)$ goes increasingly from $(0,0)$ till $(1,2)$ and $h(x)=\lfloor x^2\rfloor+1=1$ there is another solution, intersection of $f(x)$ with $h(x)$.

This second solution is not at all evident as the first one was. A way of approximation is calculation of $f(x)$ for the values $0.9,0.8,0.7$ so we get $f(0.7)\lt1\lt f(0.8)$ then certain $x$ such that $0.7\lt x\lt 0.8$ is the other solution. Follow this way we can get $0.76\lt x\lt0.77$ then we can get $0.763\lt x\lt0.764$ and so on till we stop our approximation.