# Thread: Solving for x....when x is both an exponent and denominator

1. ## Solving for x....when x is both an exponent and denominator

How do you solve this equation?

[1-(1/1.05)^30]0.0635 = [(1/1.05)^30] * [(1-(1/1.05)^(x-65)] * [30/x]

See how x is both a denominator, and an exponent?

2. Is that it?

$\displaystyle \left[1-\left(\dfrac{1}{1.05}\right)^{30}\right]0.0635 = \dfrac{30}{x}\left[\left(\dfrac{1}{1.05}\right)^{30}\right]\left[1-\left(\dfrac{1}{1.05}\right)^{x-65}\right]$

Can you simplify this more? Work out for example all the parts containing the values without x?

3. All of the numbers are irrational; they get kind of messy :/ is there a way to do it, though?

4. Originally Posted by RU091
How do you solve this equation?

[1-(1/1.05)^30]0.0635 = [(1/1.05)^30] * [(1-(1/1.05)^(x-65)] * [30/x]

See how x is both a denominator, and an exponent?
Something tells me you cannot solve this algebraically...

Moving the x's to one side and the values to the other, we get:

$\displaystyle \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} = \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x}$

Taking logarithms of both sides won't help us. The best idea may be to use an approximation method to solve for x.

5. What7s an approximation method? How do you do that?

6. An approximation method approximates x, as you cannot get the precise answer. It gets a decimal approximation down to however many decimal places of accuracy you want.

The easiest one I can think of is the Newton Raphson method, but finding the derivative of the function isn't too easy.

7. Originally Posted by RU091
How do you solve this equation?

[1-(1/1.05)^30]0.0635 = [(1/1.05)^30] * [(1-(1/1.05)^(x-65)] * [30/x]

See how x is both a denominator, and an exponent?
1. Simplify the coefficients.

2. Do you want real roots? It does not have any! If you are interested in where it changes sign then that would be at x=0, but it is undefined there.

CB

8. Originally Posted by Educated
Something tells me you cannot solve this algebraically...

Moving the x's to one side and the values to the other, we get:

$\displaystyle \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} = \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x}$

Taking logarithms of both sides won't help us. The best idea may be to use an approximation method to solve for x.
Can be solved using Lambert's W function (which won't give a real root in this case) but that would not really help the OP so I won't do that here.

CB

9. Originally Posted by CaptainBlack
2. Do you want real roots? It does not have any!
CB
Really? I got 2 real roots from the equation: $\displaystyle \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} - \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x} = 0$

10. Originally Posted by Educated
Really? I got 2 real roots from the equation: $\displaystyle \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} - \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x} = 0$

Well the Maxima simplifies (evaluates coefficients numerically because of decimal coeficients):

KK:(1-(20/21)^30)*(0.0635)/30/(20/21)^30-(1-(20/21)^(x-65))/x;

$\displaystyle 0.0070314446940689-\frac{1-{21}^{65-x}\,{20}^{x-65}}{x}$

Which now gives a root at ~=-183.5, but no other (why I now have a root I don't know but appears to be something to do with using 20/21 for 1/1.05), but I think it is wrong - more checking needed me-thinks

(Switching back to 1/1.05 takes us back to the previous situation with no roots and plots now plot what the function evaluates to

I have changed the definition of KK to functional form which gives LaTeX to compare with your's:

$\displaystyle \mathrm{KK}\left( x\right) :=\dfrac{\frac{\left( 1-{\left( \frac{1}{1.05}\right) }^{30}\right) \,0.0635}{30}}{{\left( \frac{1}{1.05}\right) }^{30}}-\dfrac{1-{\left( \frac{1}{1.05}\right) }^{x-65}}{x}$

Which looks OK to me)

CB