1. ## Equation

Find the solutions of x and n in this equation.( x^2+615=2^n)

2. ## Re: Equation

Solving for x...

\displaystyle \displaystyle \begin{align*} x^2 + 615 &= 2^n \\ x^2 &= 2^n - 615 \\ x &= \pm \sqrt{2^n - 615} \end{align*}

Solving for n...

\displaystyle \displaystyle \begin{align*} 2^n &= x^2 + 615 \\ n &= \log_2{\left(x^2 + 615\right)} \end{align*}

3. ## Re: Equation

No, I mean the definite solution. I can get one solution from graph,but there exists no other solution. How can I prove it?

4. ## Re: Equation

Hello, Swarnav!

$\displaystyle \text{Find the solutions of }x\text{ and }n\text{ in this equation: }\:x^2+615\:=\:2^n$

Are there any restrictions on $\displaystyle x$ and $\displaystyle n$?

$\displaystyle \text{Solving for }x\!:\;\;x \;=\;\pm\sqrt{2^n-615}$

$\displaystyle \text{If }x\text{ must be real: }\;2^n - 615 \:\ge\;0 \quad\Rightarrow\quad n \:\ge\:\tfrac{\ln615}{\ln2} \:=\:9.2644426...$

I don't see any difficulty . . .

$\displaystyle \text{If }x = 9.3\text{, then }x = \sqrt{15.34593963...}$

$\displaystyle \text{If }n = 10\text{, then }x = \sqrt{409}$

$\displaystyle \text{If }n = 11\text{, then }x = \sqrt{1433}$

. . $\displaystyle \text{ . . . and so on . . .}$

$\displaystyle \text{If you insist that }x\text{ and }n\text{ are both } integers,$
. . $\displaystyle \text{then we have a }big\text{ problem.}$

You say you found one solution graphically.
I suspect that you graphed: .$\displaystyle y \:=\: x^2+615\,\text{ and }\,y \,=\, 2^x$
. . and found their intersection.
In doing so, you assumed that $\displaystyle x = n.$

5. ## Re: Equation

Yes x and n are both integers. Sorry I forgot to mention it. For this reason I cannot find solution of x. But there is one solution,if n=12 then we can get x as a integer.