Need Help

[sirjr1313]

Need help solving. Not really sure where to start.

5x^2/25^x=125

 

[lookagain]

Can't be solved directly; iteration required.
Anyhow, easy to solve by inspection: x = -1

A solution is found to be x = -1, for example, by inspection.

However, there must be work/explanation to determine if there
could be any more real solutions.

 

[Deleted member 4993]

Need help solving. Not really sure where to start.

5x^2/25^x=125

5x² = 5^2x * 5³

x² = 5^2(x+1)

x = ± 5^x+1

Now you can plot y =x and y = - 5^x+1 and y = 5^x+1- and the points of intersections are your solutions.

 

[lookagain]

5x² = 5^2x * 5³

x² = 5^2(x-1)


= 5^x-1

Now you can plot y =x and y = 5^x-1 - and the points of intersections are your solutions.

The correct equation is \(\displaystyle \ \ x^2 \ = \ 5^{2(x + 1)}.\)

From there, \(\displaystyle \ x \ = \ \pm5^{x + 1}.\)

\(\displaystyle x \ = \ 5^{x + 1} \ \ has \ \ no \ \ real \ \ solutions.\)

Edit:

In a corresponding way, a graph shows that \(\displaystyle y \ \ = \ x \ \ and \ \ y \ = \ -5^{x + 1} \ \ \) cross each other once. There exists no points of tangency. That is, one does not just touch the other.