solve the equation 2^x = 200x

[mistjama]

I am having trouble solving the following equation algebraically. Can someone help?

2^x = 200x

I think you need to take the (ln) of both sides, but then I get stuck.

 

[stapel]

I don't believe that this equation can be solved algebraically. You'll have to resort to numerical methods.

 

[mistjama]

Re:

I don't believe that this equation can be solved algebraically. You'll have to resort to numerical methods.

But if you graph them each [y= 200x and y = 2^x], they meet around x= 11. What prevents an algebraic solution?

 

[Denis]

Re: Re:

But if you graph them each [y= 200x and y = 2^x], they meet around x= 11. What prevents an algebraic solution?

They meet close to x = 11.12
Whatever "program" created the graph used numerical methods, or iteration: NOT an algebraic solution.
WHY? Because (as you were told) your equation CANNOT be solved directly.
Your teacher will tell you that...though I wonder why you haven't been told already :shock:

 

[mistjama]

Sorry to be a pain. The teacher hasn't explained it to me because I am the teacher. This problem was on a state teacher exam (middle school math). They didn't ask for an exact solution. Each equation was a formula for computing employee bonuses. The problem was to figure out which formula was most beneficial to workers in which years, from x=1 to x=20.
I just got to wondering, since the lines cross, how you could find exactly WHERE they cross. The word "iteration" implies maybe calculus could solve it? I will go and ask some friends.

 

[Denis]

I just got to wondering, since the lines cross, how you could find exactly WHERE they cross. The word "iteration" implies maybe calculus could solve it? I will go and ask some friends.

Well, at least you got friends :wink:

"Iteration" is a fancy word for "guesswork"! Try googling it.

3^x = 17 ; x = log(17) / log(3) : no problem!

3^x = 17x ; x = log(17x) / log(3) : oh oh !
x^x = 17 ; x = log(17) / log(x) : oh oh !