Solve x^2+2^x=100: Step-by-Step Guide for Clear Understanding

[doc102]

Thanks... Please show step. i am kinda confused

 

[robphy]

http://www.wolframalpha.com/input/?i="x^2++2^x+=+100+"

[edit: My reply above was as terse as the original version of your question.
Can you show some work on how you approached the problem?]

 

[Mentallic]

Thanks... Please show step. i am kinda confused

These sorts of functions don't have any simple solutions, but you can always find good approximations to them using various numerical techniques.

For this particular equation though, it just so happens that x=6 satisfies. Also, since x=-10 gives us (-10)²+2^-10=100+2^-10, which is just 1/1024 larger than 100 so then this other solution must be a tiny bit more positive than -10.

How do we know that there are only two real solutions? For x<0, x² is the dominant term as 2^x is only a smaller value between 0 and 1, so as x² grows large as x gets larger in the negative direction, it will eventually cross the line y=100 and never come back. For x>0, both terms are dominant and grow larger, hence it must cross y=100 there too and also never come back. Hence we only have two solutions.

 

[pwsnafu]

This particular problem comes up in math quizes, and usually they ask for the positive solution. Clearly the solution is less than 10.
Suppose that the solution is a rational, $x=p/q$. Then
$\frac{p^2}{q^2} + 2^{p/q} = 100$.
The first term is rational, the right hand side is an integer. What does this tell you about the solution?

 

[Maths Absorber]

I guess you can use hit and trial to find out that x = 6 satisfies this equation. If it's known that the equation has integral roots, it's easy but otherwise ...
Maybe logarithms would be useful

 

[abbas_majidi]

Your problem have numerical solutions and they are attractive. See below picture which it is calculated with Mathematica:

This equation have two answer and both of them are very near to integer numbers. I tried to find analytical solution but I can't.

 

[Murtuza Tipu]

You could try trial and error method. As we have 2^x and x^2 in equation , x^2 is always positive hence 2^x is less than 100. hence x is less than 6,now put values of x=1,2,3,4,5,6. Here x=6 satisfies equation .Hence soln. is x=6.