Can "2 ^ n = 2 * n" be solved?

Question

Is there a way, other than by inspection or graphically, of mathematically showing the solutions for n?

Answer 1

I see what you're saying, and it looks like these other answers aren't doing what you seek.  You can accomplish this mathematically using logs (natural logs I'll use).

2^n = 2*n  

ln(2^n) = ln(2*n)  (took the log of both sides)

ln(2)*n = ln(2*n) = ln(2) + ln(n) (applied log characteristics)

ln(2)*n-ln(2) = ln(n) (subtracted ln(2) from each side)

ln(2)*(n-1) = ln(n) (applied distributive property)

e^(ln(2) * (n-1)) = e^(ln(n))

(2^n)/2 = n  which goes right back to:

2^n = 2*n

WTF.  Please message me when u figure this out.

Answer 2

http://en.wikipedia.org/wiki/Transcenden...

A function 2ⁿ (or any function cⁿ, n is a variable) is a transcendental function. By definition of what transcendental functions are, they are not algebraic. Since there is the algebraic part on the right and the transcendental part on the left, there is no algebraic way to combine them for a solution.

There are many techniques of analysis that can eliminate ranges of values (like by observing the exponentials are always positive so negative values of n will not work), provide certainty of the results (like by saying that they are both increasing but one is increasing at a constant rate, thus giving a maximum of two solutions), and manipulate them into easier to guess forms (though the form it is already in what is likely the easiest).

There is no simple way to just find a solution for problems of this form though. You can take something like Newton's method and seek out zeroes and find them quite quickly (especially with a calculator or math program). That will give you better and better approximations (and eventually you would actually realize, "Hey, it's one." for simple values).

http://en.wikipedia.org/wiki/Newton%27s_...

That's a useful way of solving such problems.

Informed guessing is actually a time-honored technique of mathematics and science though, no matter what anyone tells you.

----

If you haven't been scared off by that much, note that you can actually use the Lambert function for a problem of this nature:

http://en.wikipedia.org/wiki/Lambert_W_F...

(There is an similar problem worked out at "Example 1")

Answer 3

There isn't a way to "solve" for n in the normal sense of using algebra to find solutions.

I suppose a proof would be to show that it works for n = 1, n =2, and is impossible for

case 1: all n's <1,

case 2: between 1 and 2

case 3: n's >2.

You may even have to break down the subcase of between 0 and 1.

For example, case 3,  prove that 2^n > 2n for all n>2.   Pretty easy to see by way of just putting in some numbers, and case 2 actually follows from this as well. Last, prove case 1. I'm too tired to think about that right now, but it won't work either (if you rewrite as 2^n-2n=0 then you'll see the left side ends up positive all the time). This might even all be the same case. I'm not sure how a rock solid mathematical proof would look like for this, but I'm pretty sure I'm on the right track about it.

Answer 4

yes it can be

if n=2

putting the value in equation 2 ^ n = 2 * n

2^2 = 2*2

4=4

hence proved :)  

Answer 5

Yes because n=2

2^2=2*2=4

soooooooo yeah.

Answer 6

n*n = 2*n

divide both sides by n

n = 2

Answer 7

yes it can be solved

2^n = 2*n  

substitute values from 0  

for n=0

it becomes

1=0   not possible

try n=1

2=2 so n=1 is one possibility

try n=2

4=4 so n=2 is another possibility  

if u dnt like this trial and error method u can take logarithm on both sides to solve it

good luck

Answer 8

Yes.

Graph in Y1=2^n

Graph in Y2=2n

The exponential function and the linear function pass each other. Use the table function on the calculator.