Solving a constant as a function of x

[How]

Hi everyone,

Appreciate if anyone can simplify/rearrange to obtain a as a function of x

(1-a)^x=0.

My solution is as below, can anyone check if this is correct?

e^0=e^(1-a)^x
1=e^(1-a)^x
1=x*e^(1-a)
ln(1/x)=(1-a)
a=1-ln(1/x)

 

[Deleted member 4993]

Hi everyone,

Appreciate if anyone can simplify/rearrange to obtain a as a function of x

(1-a)^x=0.

My solution is as below, can anyone check if this is correct?

e^0=e^(1-a)^x
1=e^(1-a)^x
1=x*e^(1-a)
ln(1/x)=(1-a)
a=1-ln(1/x)

e^0=e^[(1-a)^x]
1=e^[(1-a)^x]
1=x*e^(1-a) ..................................Incorrect


ln(1/x)=(1-a)
a=1-ln(1/x)

For example:

\(\displaystyle 2^{3^2} = 2^9 = 512\)

It is NOT EQUAL to:

\(\displaystyle 2^{3^2} \ne 2* 2^3 \)

 

[JeffM]

This is one of those problems where you do not need techniques; rather you need to think about fundamentals.

[MATH]a \ne 1 \implies 1 - a \ne 0 \implies (1 - a)^x = 0 \text { under what circumstanes.}[/MATH]

 

[Cubist]

This is one of those problems where you do not need techniques; rather you need to think about fundamentals.

[MATH]a \ne 1 \implies 1 - a \ne 0 \implies (1 - a)^x = 0 \text { under what circumstanes.}[/MATH]

Under no circumstances I think, surely no value of x could make that true? But you are thinking along the same lines as me...

OP, do you know what \(\displaystyle y^x=0 \) does imply about y (and it also implies something about x) ?

If so then you can simply state \(\displaystyle y^x=0 \implies y=...\)

 

[Cubist]

Under no circumstances I think, surely no value of x could make that true?

Actually there is a value of x (and a range of "a" values) that satisfies JeffM's suggestion if we consider \(\displaystyle \lim_{x \to ?}(1-a)^x = 0\) but there's a simpler solution that does not involve limits.

 

[How]

Actually there is a value of x (and a range of "a" values) that satisfies JeffM's suggestion if we consider \(\displaystyle \lim_{x \to ?}(1-a)^x = 0\) but there's a simpler solution that does not involve limits.

It will be awesome if there's a simpler solution to this... my background is engineering so I only know how to solve the basic ones

 

[Cubist]

It will be awesome if there's a simpler solution to this... my background is engineering so I only know how to solve the basic ones

What number, when squared, is zero?

 

[Deleted member 4993]

What number, when squared, is zero?

Multiplying by zero is illegal - even in the name of squaring!!!

 

[Cubist]

Multiplying by zero is illegal - even in the name of squaring!!!

Really

Please explain, I've not heard of this. My calculator will multiply by zero

 

[pka]

Multiplying by zero is illegal - even in the name of squaring!!!

Why would you write that?
This is theorem 1 in my ring theory notes.
\(\displaystyle \forall a\in\mathscr{R}[~a\cdot 0=0=0\cdot a]\) clearly that includes \(\displaystyle 0\cdot 0\)

 

[JeffM]

Actually there is a value of x (and a range of "a" values) that satisfies JeffM's suggestion if we consider \(\displaystyle \lim_{x \to ?}(1-a)^x = 0\) but there's a simpler solution that does not involve limits.

Actually, you are ascribing to me a far more sophisticated thought than I was trying to elicit.

I had no thought about limits.

What number can I raise to a power and get zero. There is such a number if we exclude the power of zero.

 

[Steven G]

Hi everyone,

Appreciate if anyone can simplify/rearrange to obtain a as a function of x

(1-a)^x=0.

My solution is as below, can anyone check if this is correct?

e^0=e^(1-a)^x
1=e^(1-a)^x
1=x*e^(1-a)
ln(1/x)=(1-a)
a=1-ln(1/x)

If x>0 then (1-a)^x= 0 only when 1-a=0 or a=1.
If x<0, then there is no solution.
So, the solution is a(x)=1

 

[Cubist]

OP, did this question come from a real world engineering problem? I ask because you might be interested in the circumstances when (1-a)^x is close to zero (which would then imply a relationship between a and x).

 

[HallsofIvy]

I think Subhotosh Kahn has been smoking the funny weed.

 

[Mr. Bland]

Appreciate if anyone can simplify/rearrange to obtain a as a function of x

(1-a)^x=0.

So [MATH]a = f(x)[/MATH], solve for [MATH]f[/MATH]?

Let's see...

[MATH](1 - a)^x = 0[/MATH]
[MATH]1 - a = \sqrt[x]{0}[/MATH]
[MATH]a = 1 - \sqrt[x]{0}[/MATH]​

Houston, we have a problem.

 

[HallsofIvy]

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.

 

[Mr. Bland]

[... T]here is NO x such that \(\displaystyle (1- a)^x= 0\).

In general, this is true. If [MATH]a = 1[/MATH] specifically, then I believe [MATH]x > 0[/MATH] will work.

 

[Deleted member 4993]

Why would you write that?
This is theorem 1 in my ring theory notes.
\(\displaystyle \forall a\in\mathscr{R}[~a\cdot 0=0=0\cdot a]\) clearly that includes \(\displaystyle 0\cdot 0\)

Really

Please explain, I've not heard of this. My calculator will multiply by zero

I could not find an emoji with "tongue-implanted-in-cheek". I thought the reference to"legality" would be frivolous enough.....

 

[JeffM]

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.

There is no need to solve this with logs.

I cannot believe how many are refusing to admit that

[MATH]x > 0 \text { and } a = 1 \implies (1 - a)^x = 0^x = 0 [/MATH]

 

[Steven G]

There is no need to solve this with logs.

I cannot believe how many are refusing to admit that

[MATH]x > 0 \text { and } a = 1 \implies (1 - a)^x = 0^x = 0 [/MATH]

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