How to solve the following equation with exponents for x : 1-(1-x/y)^x = 1/2

[Tigrou]

Hello. I'm trying to solve the following equation for x :

\(\displaystyle 1-\left(1-\dfrac{x}{y}\right)^x = \dfrac{1}{2}\)

I put all unknown at the left side :

\(\displaystyle \left(1-\dfrac{x}{y}\right)^x = 1-\dfrac{1}{2}\)

\(\displaystyle \left(1-\dfrac{x}{y}\right)^x = \dfrac{1}{2}\)

Then I use the following property :

\(\displaystyle a ^ x = b \Leftrightarrow x =\dfrac{ ln\left(b\right) }{ln\left(a\right)} \)

\(\displaystyle x = \dfrac{ ln\left(\dfrac{1}{2}\right) }{ ln\left(1-\dfrac{x}{y}\right) } \)

I got this but then i'm stuck :

\(\displaystyle x \cdot ln\left(1-\dfrac{x}{y}\right) = ln\left(\dfrac{1}{2}\right) \)


I have tried is to go further but i'm not sure this is the right way :

\(\displaystyle x \cdot ln\left(\dfrac{y}{y}-\dfrac{x}{y}\right) = ln\left(\dfrac{1}{2}\right) \)

\(\displaystyle x \cdot ln\left(\dfrac{y-x}{y}\right) = ln\left(\dfrac{1}{2}\right) \)

\(\displaystyle x \cdot \left(ln\left(y-x\right) - ln\left(y\right)\right) = ln\left(\dfrac{1}{2}\right) \)

\(\displaystyle x \cdot ln\left(y-x\right) - x \cdot ln\left(y\right) = ln\left(\dfrac{1}{2}\right) \)

 

[ksdhart2]

First off, a big thank you for showing us your efforts! Lately it seems like we've been inundated with people who either didn't read the rules at all, or else just can't be bothered to follow them. So, it's refreshing to see someone who did.

That aside, to the problem at hand. Everything up to this step is good:

\(\displaystyle x * ln \left( 1-\dfrac{x}{y} \right) = ln \left( \dfrac{1}{2} \right)\)

At this point, a good next step would be to divide both sides by x:

\(\displaystyle ln \left( 1-\dfrac{x}{y} \right) = \dfrac{ln \left( \frac{1}{2} \right)}{x}\)

Then raise e to the power of both sides (as well as rewrite ln(1/2) as -ln(2)):

\(\displaystyle 1-\dfrac{x}{y} = e^{\frac{-ln(2)}{x}}\)

This then simplifies down to

\(\displaystyle 1-\dfrac{x}{y} = 2^{-\frac{1}{x}}\)

Now you try continuing from here. See where this leads you.

 

[Tigrou]

Thanks for you help.

Could you give me more info about which property you used when raising e to the power (eg : a link to free math help website or wiki)

Also I tried to go further with your last step.

However I'm still stuck. If I use this : \(\displaystyle a ^ x = b \Leftrightarrow x =\dfrac{ ln\left(b\right) }{ln\left(a\right)} \) I feel like i'm going back to what I got before.

Can you give me some hints about how to isolate x ?

 

[Deleted member 4993]

Thanks for you help.

Could you give me more info about which property you used when raising e to the power (eg : a link to free math help website or wiki)

Also I tried to go further with your last step.

However I'm still stuck. If I use this : \(\displaystyle a ^ x = b \Leftrightarrow x =\dfrac{ ln\left(b\right) }{ln\left(a\right)} \) I feel like i'm going back to what I got before.

Can you give me some hints about how to isolate x ?

\(\displaystyle 1-\dfrac{x}{y} = 2^{-\frac{1}{x}}\)

\(\displaystyle \dfrac{x}{y} = 1 - 2^{-\frac{1}{x}}\)

\(\displaystyle y = \dfrac{x}{1 - 2^{-\frac{1}{x}}}\)

Continue.......

 

[ksdhart2]

Thanks for you help.

Could you give me more info about which property you used when raising e to the power (eg : a link to free math help website or wiki)

Certainly. Powers involving e can often seem mystical and magical, but they're really not. They just involve using the properties of exponents and the properties of exponents and logarithms. In this specific case, we can use this rule:

\(\displaystyle a^{\frac{b}{c}} = \left( a^b \right)^{\frac{1}{c}}\)

Applying this rule to the given fraction leaves:

\(\displaystyle e^{\frac{-ln(2)}{x}} = \left( e^{-ln(2)} \right)^{\frac{1}{x}}\)

Next, use the fact that \(\displaystyle e^x\) and \(\displaystyle ln(x)\) are inverses:

\(\displaystyle \left( e^{-ln(2)} \right)^{\frac{1}{x}} = \left( 2^{-1} \right)^{\frac{1}{x}}\)

Recombining the exponents gives the final answer:

\(\displaystyle \left( 2^{-1} \right)^{\frac{1}{x}} = 2^{-\frac{1}{x}}\)

Et voila.

 

[tkhunny]

It is one of the great disappointments of the early study of mathematics that some things cannot be solved conveniently.

\(\displaystyle x = e^{x}\) may be the simplest example.

 

[Tigrou]

I'm still struggling with Subhotosh Khan las step :

\(\displaystyle y = \dfrac{x}{1 - 2^{-\frac{1}{x}}}\)

If multiply both sides by \(\displaystyle 1 - 2^{-\frac{1}{x}}\) I mix \(\displaystyle x\) and \(\displaystyle y\) again.

Is there a way to simplify \(\displaystyle 2^{-\frac{1}{x}}\) (considering it's inside the fraction ?)

I have drawn a graph from the equation (and swap x and y), it looks like a parabola (2nd degree equation)

 

[mmm4444bot]

\(\displaystyle y = \dfrac{x}{1 - 2^{-\frac{1}{x}}}\)

… I have drawn a graph from the equation (and swap x and y), it looks like a parabola …

Graphing is one way, to visualize the inverse of the function above, but look closer -- in the vicinity of the Origin -- and you'll discover that the graph is not a parabola.

Is there a way to simplify \(\displaystyle 2^{-\frac{1}{x}}\)

There are a number of ways to express 2^(-1/x). Yet, rewriting that expression will not help to find an algebraic solution for x because no basic functions of y exist for x, in the given relationship between x and y.

From where does this exercise come? :cool:

 

[ksdhart2]

There are a number of ways to express 2^(-1/x). Yet, rewriting that expression will not help to find an algebraic solution for x because no basic functions of y exist for x, in the given relationship between x and y.

Oof! Can't believe I didn't notice this myself. I didn't real carefully enough I suppose. I had thought they were asking for a solution of the form y = (expression in x), which, as demonstrated is totally do-able. Yet, the opposite, x = (expression in y) doesn't seem possible given only the elementary functions. Indeed, even Mathematica, when asked to calculate:

Solve[1-(1-x/y)^x==1/2,x]

fails to solve it, noting that "This system cannot be solved with the methods available to Solve"

 

[Tigrou]

From where does this exercise come? :cool:

It's not an exercise. It's from me

x is basically the number of times you have to pick up random one element inside a list of y elements to have exactly 50% of probability choosing one element which is inside the x first elements.

That probability is given by :

\(\displaystyle 1-\left(1-\dfrac{x}{y}\right)^x\)

Let's say we have 4 elements. We pickup a random element 3 times (and put it back each try).
What is the probability to pickup the first one at least once ?

\(\displaystyle 1-\left(1-\dfrac{1}{4}\right)^3\) = 58% of probability

Indeed, even Mathematica ... fails to solve it, noting that "This system cannot be solved with the methods available to Solve"

I think it's solvable

If I wrote this in Wolfram (I fixed the y value) :

solve (1 - (1 - (x/10000)) ^ x = 1/2)

It will find x = 83.0821 which is right.
It means you have to pickup 83 numbers randomly to have 50% of probability choosing one amongst the first 83 numbers.

Unfortunately Wolfram doesn't want to show the detail (it require Pro version)

 

[tkhunny]

There are many ways to track down numerical solutions when no algebraic solution can be had.

Bisection
Newton's Method

Just to name a couple.

It is NOT a matter of trail and error, but one does just have to be lucky, sometimes.

 

[mmm4444bot]

I think it's solvable

This "it" refers to something other than what you asked for in your OP.

Your original equation is not solvable for x, with familiar algebraic steps.

If I wrote this in Wolfram (I fixed the y value) :

solve (1 - (1 - (x/10000)) ^ x = 1/2)

It will find x = 83.0821 which is right.

That is an approximation for x, but you didn't ask for an approximation.

Picking a value for y and then calculating the corresponding x value is easy. That is not the same as finding a function (i.e., expression) for x in terms of y.

When I input your original question at Wolfram, they provide an implicit plot and state that x cannot be expressed in terms of known functions of y.

 

[Tigrou]

@mmm4444bot : ok got it. So basically this equation is not solvable. We can only approximate x and that's it. y can be solved btw.


I have tried the whole weekend to solve it but couldn't now i know why.
I did not know there was unsolvable equations. At least not ones that looks so simple.

 

[tkhunny]

No, it is quite "solvable", just not with pretty, closed algebraic expressions. Solutions, if any, are not at all difficult to locate. Series Methods might be interesting. They aren't finite.

 

[mmm4444bot]

so basically this equation is not solvable

Not by using the usual algebraic steps you've learned so far, with basic, familiar functions.

The reason has to do with the fact that the variable we're solving for appears as both a base and an exponent. When we take logs, to get x out the exponent position, it traps the other x inside the log function (with y). To get that x out of the logarithm (to separate it from y), we need to exponentiate. But, exponentiation puts the first x back into the exponent position. It's a Catch-22 situation.

Perhaps, one day, you will invent some new mathematics, allowing algebra students everywhere to find function x in terms of y.