Solve Exponential Equation: 5^(sqrt(x)) + (1/5)^(sqrt(x)) = 25

  • Thread starter Thread starter scientifico
  • Start date Start date
  • Tags Tags
    Exponential
AI Thread Summary
The equation 5^(sqrt(x)) + 5 * 5^(-sqrt(x)) = 25 + 1/5 can be simplified by substituting u = 5^(sqrt(x)), leading to 5^(-sqrt(x)) being expressed as 1/u. This transformation allows for easier manipulation of the equation. The next step involves solving for u, which simplifies the overall problem. The discussion emphasizes the importance of substitution in solving exponential equations effectively.
scientifico
Messages
181
Reaction score
0

Homework Statement


5^( \sqrt(x) ) + 5 * 5^(- \sqrt(x) ) = 25 + 1/5

2. The attempt at a solution

I have thought to transform 5^ (-\sqrt(x) ) in (1/5)^(\sqrt(x) ) but I don't know how to solve then... could you help me ?

thank you
 
Physics news on Phys.org
scientifico said:

Homework Statement


5^( \sqrt(x) ) + 5 * 5^(- \sqrt(x) ) = 25 + 1/5

2. The attempt at a solution

I have thought to transform 5^ (-\sqrt(x) ) in (1/5)^(\sqrt(x) ) but I don't know how to solve then... could you help me ?

thank you

That's a good idea. So if u=5^(sqrt(x)) then 5^(-sqrt(x)) equals 1/u. Solve for u first.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top