Solving a constant as a function of x

[Steven G]

You need to "bring down" the exponent, x. Just "moving it to the other side" by taking the "x"th root doesn't help. To "bring down" an exponent, you need to use a logarithm. IF the problem were, say, $\displaystyle (1- a)^x= 2$ we could say that $\displaystyle log(1- a)^x= x log(1- a)= log(2)$ so $\displaystyle x= \frac{log(2)}{log(1- a)}$. The base of the logarithm doesn't matter since it will cancel in division.

However, here, we do have a problem! Log(0) does not exist- the domain of the logarithm function is "all positive numbers" and does not include 0. What that means is that there is NO x such that $\displaystyle (1- a)^x= 0$. There is no solution to this equation.

The OP wants a as a function of x. If x>0, then a=1. If a<0, then there is no x. So in the end, a(x)=1 if a>0 and undefined elsewhere.

 

[JeffM]

So isn't the answer to OP a(x) = 1 if x>0. Where is the problem?

Look at post number 4, which explicitly refers to my post number 3. Post 4 says no value of x could make the proposition true if a is not equal to 1. Why were there any further posts after that? Obiously, a = 1 is the only possible answer (subject to x being positive).

Post number 5 seems to imply that we need limits to solve this problem.

Post number 8: "multiplying by zero is illegal" is a proposition unlikely to help the OP.

Post number 13, which again seeks to use limits to answer a very obvious question.

Post number 15, which seems to assert that 0 has no roots.

Post number 16, which asserts that the only way to solve the problem is to use logs, which is impossible, thereby implying the problem is insoluble.

 

[Steven G]

Look at post number 4, which explicitly refers to my post number 3. Post 4 says no value of x could make the proposition true if a is not equal to 1. Why were there any further posts after that? Obiously, a = 1 is the only possible answer (subject to x being positive).

Post number 5 seems to imply that we need limits to solve this problem.

Post number 8: "multiplying by zero is illegal" is a proposition unlikely to help the OP.

Post number 13, which again seeks to use limits to answer a very obvious question.

Post number 15, which seems to assert that 0 has no roots.

Post number 16, which asserts that the only way to solve the problem is to use logs, which is impossible, thereby implying the problem is insoluble.

I obviously agree with you. I can not believe the posts made in this thread.

 

[Otis]

… in the end, a(x)=1 if a>0 and undefined elsewhere.

I think you mean x>0

Then is that 'The End'?

 

[pka]

I think this rivals any as being a truly obtuse thread.
Is this a true proposition?
If $\displaystyle \left(\frac{0}{0}=1\right)\text{ then }1\ge 2~.$
All anyone needs answer is $\displaystyle T\text{ or }F$.