solve 2^x =x^2 I need steps?

Question

I know the answer is 2,4. But I need steps. I never expect any guesses. show me the steps plz.

Answer 1

I've been going around and around on this one.

By graphing:
http://img522.imageshack.us/img522/6114/x22xly0.gif

It is easy to see there are 3 solutions. And we can subtract the two and see the zeroes. Certainly calculus could be employed to take the 2nd derivative to find when the slope changes from concave to convex... etc. etc.

Or we can use methods to estimate the answers, but that doesn't seem very satisfying. It seems like there must be a way to do this without resorting to calculus and guessing.

It is easy enough to take the log[base 2] of both sides:
x = log[2] x²

Using the rule of logs (log x^k = k log x) we can get:
x = 2 log[2] x

And the change of base rule (log[2] x = log(x) / log(2) ) results in:
x = 2 log(x) / log(2)

This can be rewritten as:
log(x) / x = log(2) / 2

Well, by inspection 2 is a solution...but that's not really too exciting...

Edit:

I asked a math major friend of mine and he had this to say,

"Actually, there's no analytic way to solve this problem. In the 1700s, Johann Lambert decided to create a function that we now call the Lambert Function. If f(x) = x e^x, then the Lambert Function is the inverse of that. There are tables and tables of those values, though now they are easily calculated using infinite series.

So actually in this case there isn't a way to solve it, other than knowing what to call the function or how to calculate the inverse of an infinite series."

Answer 2

The first step is to complete the square... you can do this by taking half of the second term and then subtracting or adding from the third term to complete the square... In this case, half of 2 is 1 and so to get 1 you have to subtract 2 but then add 2 to keep everything balanced... x^2 + 2x + 3 - 2 = -2 x^2 + 2x + 1 = -2 (x + 1)^2 = -2 And there you go :D

Answer 3

This type of problem is best solved numerically.

Use Newton's method:
x_(n+1) = x_n - f(x_n)/f'(x_n)
where the sequence x_n for n=1,2,...,∞ converges to the root.

Let f(x) = 2^x - x^2

Then
f'(x) = ln(2)*(2^x) - 2x

Let your first guess be 1.
x_2 = 1 + f(1) / f'(1) = 1 - (2-1)/(2ln(2) - 2)
= (2ln(2) - 3) / (2ln(2) - 2)
= 2.6294
So this is your next guess.

...and so on.

You can do this by hand, but it might take a few iterations to start seeing convergence.

If you use a little computer program to run this algorithm (e-mail me if you want some code), you will find that, depending on your initial guess, you will have convergence to either -0.7667, 2, or 4. All three answers check.

Answer 4

let f(x)=2^x and

g(x) =x^2
and
h(x) = f(x)-g(x)= 2^x-x^2

Then h'(x) = 2^x ln 2 - 2x
h''(x)=2^x (ln 2)^2 -2

You have to show that on interval (0,2) and (4, infinity)
h> 0
and on (2,4) h<0
That will result from noticing that
1)h'' has an unique root betwen 0 and 4, that changes the graph from concave down to concave up.,
that is c=2 ln(ln 2)/(1-ln 2) approximatively 1.2
2)h has clearly roots 2 and 4.
3)h(0)=1>0

hence h goes up, then down, reaches c(changes the convexity) then down passes 2, and then up again , passes 4 and up to infinity. We notice that h changes its monotony somewhere between 2 and 4 and there is one of the roots of h' located. The other root of h' I think is between -infinity and c.

If by absurd, there is other root, h should change once again its convexity but we see from h'' that h changes its convexity only once.
In other words if a continuos function
has 3 roots, it changes the convexity at least once
if it has 4 roots, it changes the convexity at least twice
if it has 5 roots it changes the convexity at least three times and so on
Just try to make a graph and convince yourself.

notice that lim x--> - infinity h(x) = - infinity
so indeed it has another negative root

Answer 5

EDIT: kamrutha p below has a very interesting part solution. After he got his:

2^(1/2) = x^(1/x),

he claims that the answer is x=2 but unfortunately that is only one of the solutions. Like other answerers have said, x=4 and x=-0.7666.. are also the other possible answers. Unfortunately, no proper solution from me. Don't know how to do.

Answer 6

The interesting solution x=-0.7666... can be expressed as an infinite towering exponential:

-(a^(a^(a^(a^...

where a=1/sqrt(2)~.7071067810

A partial evaluation of this construction would be, for example,
-(a^(a^(a^(a)))). In the limit, as the length of the partial construction grows to infinity, its value will converge to the solution x=-0.7666...

Answer 7

Take natural log of each side
ln2^x = lnx^2
Law of logs says you can move exponent in front
xln2 = 2lnx
get variable on same side
x/lnx = 2/ln2
after that im lost sorry

Answer 8

solve 2^x =x^2

I found 3 zeros

x1= -0.7666
x2 = 2
x3 = 4

I am working on the analytical solution

Answer 9

it can solved by means of logarithms

2^x =x^2
let 2^x = x^2
taking logarithm on both the sides
log(2^x) = log( x^2)
xlog 2 = 2log x
log2 / 2 = logx / x
log 2^0.5 = log x^ 1/x
since log is there on both the sides
2^1/2 = x^1/x
hence x = 2

Answer 10

2^x = x^2 is not right because if x is 5 then 5^2 is 25 and 2^5 is 32 so your wrong.