exponential equation?? (1 Viewer)

[akuchan]

2^x = x + 1

solve for x

=/ any ideas? i know x = 0 x = 1
but dunno the actual working out for it

 

[watatank]

Draw a graph

where y = 2^x meets y = x + 1

 

[jyu]

2^x = x + 1

solve for x

=/ any ideas? i know x = 0 x = 1
but dunno the actual working out for it

I don't think it can be solved analytically (algebraically). If you look at the two graphs, they intersect at two points only around x = 0 and x = 1. These two values happen to satisfy the equation, so they are the only solutions. Use numerical methods for other similar equations.

:wave:

 

[pLuvia]

You can use the Newton's method or bisection method

 

[watatank]

You can use the Newton's method or bisection method

Can you? Aren't those approximation methods?

 

[pLuvia]

Their just approximation methods to find roots for equations, Newton's method uses derivatives and the bisection method halves the values you get. You should learn it later one, but they are not that important as they rarely in exams.

This method is basically like the graphing approximation method, but this is much more accurate (Newton's method > Bisection method) unless you have a VERY accurate graph then you can see where they intersect.

But with this question if you want to use either of these bisection you should move all values to one side, makes it easier