Real Solutions to Equations [2^x = 1+x^2 & 3^x+4^x+5^x = x^2]

[juantheron]

No. of real solution of the equations

[1]:: $2^x = 1+x^2$

[2]:: $3^x+4^x+5^x = x^2$

My Trail for solution::

[1] Let $f(x) = 2^x-1-x^2$. Here Clearly $x = 0$ and $x = 1$ are solution of Given equation. Now We will find any other solution exists OR not.

So we use Derivative Test

$f^{'}(x) = 2^x \cdot \ln(2) - 2x$ and $f^{''}(x) = 2^x \cdot (\ln(2))^2-2$ and $f^{'''}(x) = 2^x \cdot (\ln(2))^3 > 0\forall x\in \mathbb{R}$

So Here $f^{'}(x)$ is Concave Upward.(Means Max. at Infty. but no idea about min.)

So $f^{'}(x) $ has at most $2$ real solution.

Now I did not understand it after that

Help Required.

Thanks

 

[Prove It]

No. of real solution of the equations

[1]:: $2^x = 1+x^2$

[2]:: $3^x+4^x+5^x = x^2$

My Trail for solution::

[1] Let $f(x) = 2^x-1-x^2$. Here Clearly $x = 0$ and $x = 1$ are solution of Given equation. Now We will find any other solution exists OR not.

So we use Derivative Test

$f^{'}(x) = 2^x \cdot \ln(2) - 2x$ and $f^{''}(x) = 2^x \cdot (\ln(2))^2-2$ and $f^{'''}(x) = 2^x \cdot (\ln(2))^3 > 0\forall x\in \mathbb{R}$

So Here $f^{'}(x)$ is Concave Upward.(Means Max. at Infty. but no idea about min.)

So $f^{'}(x) $ has at most $2$ real solution.

Now I did not understand it after that

Help Required.

Thanks

For the first one, just by inspection we can see that $$\displaystyle \begin{align*} x = 1 \end{align*}$$ is a solution.

Now notice that if $$\displaystyle \begin{align*} f(x) = 1 + x^2 + 2^x \end{align*}$$, then $$\displaystyle \begin{align*} f'(x) = 2x + 2^x\ln{(2)} \end{align*}$$, which is obviously positive if $$\displaystyle \begin{align*} x > 0 \end{align*}$$, and since the function is increasing it can never get back to 0.

Also note that $$\displaystyle \begin{align*} 1 + x^2 \geq 1 \end{align*}$$ for all x, while $$\displaystyle \begin{align*} 2^x < 1 \end{align*}$$ for all $$\displaystyle \begin{align*} x < 0 \end{align*}$$, so there can't be any solutions if $$\displaystyle \begin{align*} x < 0 \end{align*}$$.

So that means there is only one solution, and it happens to be x = 1.

 

[topsquark]

[1]:: $2^x = 1+x^2$

So that means there is only one solution, and it happens to be x = 1.

Actually there are three real solutions...

-Dan

Edit: I should mention that the x > 4 solution has to be done numerically.

 

[Prove It]

Oops, my derivative was incorrect, sorry :(

 

[topsquark]

Oops, my derivative was incorrect, sorry :(

It wasn't the derivative, but f(x) = 2^x - 1 - x^2. You said it was f(x) = 2^x + 1 + x^2. Easy enough to make the mistake.

-Dan