0
$\begingroup$

I'm in no way unfamiliar with these sorts of equations and have had plenty of practice with them, however I'm going insane over a question I feel I should be able to easily answer algebraically but am stuck on. This is the question:

Solve for $x$ & $y$.

1: $2^x = y$

2: $3-x = y$

Sorry for boring you with such a mundane question, but its driving me mental.

Thanks for any help.

(Sorry if there are any formatting issues, I'm new to this StackExchange).

  • 0
    Hint: have you tried to eliminate $y$?2017-01-08
  • 0
    Eliminating $y$ you get the equation $2^x=3-x$.2017-01-08
  • 0
    Putting all the $x$s on one side, $x+2^x=3$.2017-01-08
  • 0
    Since the function $x+2^x$ is strictly increasing, there can't be more than one solution.2017-01-08
  • 0
    Thanks for all the solutions! I realised how to eliminate y, I was more concerned with solving 2^x = 3−x. I never knew what a transcendental function was, so thanks to everyone for mentioning it.2017-01-08

3 Answers 3

1

We can easily see these equations are satisfied for $x=1,y=2$.

To prove that it is the only solution.

$\ \ \ \ \ 3-x=2^x$

$\Rightarrow 3=x+2^x$

$x+2^x$ is monotonically increasing function. Hence $Y=3$ intersects $Y=x+2^x$ curve at only one-point.

1

For the equation, you have to solve $2^x=3-x.$

For this kind of transcendental equation, we may not have a standard way to solve it. My hint on this is, it has an obvious solution $x=1,$ and you may be able to argue that if this equation only have one solution. Maybe by drawing the graph, and give some discussions.

1

Since both RHS = y is equivalent to the single equation. $$ 2^x = 3-x. $$

If you haven't found a way to solve this algebraically, I don't blame yourself. There is no way to 'solve for x' in algebra. This is a transcendental function and must be solved graphically.

If you graph $2^x$ and $3-x$, you'll see they intersect once. By inspection $x =1$ is a solution, so that's the one real solution.

(additionally there are complex solutions in terms of the Lambert W-function)