How to solve this equation?

peterfede

New member
Joined
Oct 2, 2020
Messages
9
Hi everyone,
could you tell me, step-by-step, how to solve the following equation:

s*x^(alpha) = (n + g + sigma)*x

I have to solve for x, but my main issue is the alpha over the x.

Thanks for your help
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,405
Hi everyone,
could you tell me, step-by-step, how to solve the following equation:

s*x^(alpha) = (n + g + sigma)*x

I have to solve for x, but my main issue is the alpha over the x.

Thanks for your help
s*x^(α) = (n + g + σ) * x

Assuming x\(\displaystyle \ne \)0, divide both sides of te equation by x and

continue......

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 

peterfede

New member
Joined
Oct 2, 2020
Messages
9
Thanks Mr Khan for your response. Let me more precise, the ending result shall be in function of x.
x can't be cancelled out. I don't know how to write alpha off from x. This is my issue.
I have the ending solution but I can't reach it. It is :
x = [s/(n+g+σ )]^(1/(1-α ))
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,405
Thanks Mr Khan for your response. Let me more precise, the ending result shall be in function of x.
x can't be cancelled out. I don't know how to write alpha off from x. This is my issue.
I have the ending solution but I can't reach it. It is :
x = [s/(n+g+σ )]^(1/(1-α ))
You have NOT shown any work done by you yet.

It is only three more steps from:

s*x^(α) = (n + g + σ) * x

I'll climb one more step and you figure the next steps:

s*x^(α-1) = (n + g + σ)
 

peterfede

New member
Joined
Oct 2, 2020
Messages
9
Actually this equation is the result of previous computations and it make sense in the Solow model. I was just stuck at the ending point. I appreciate your response.
So, if I'm right, on the right side you brought x at the denominator, added -1 as exponent and then multiplied for the left side. By exploiting the power rule, you added -1 to the power of x.
At this stage I'd do as follows:
Raise both left and right side by 1/(α-1).
So, x= ((n + g + σ)/s) ^1/(α-1).
Not reached the solution yet. Can I have another additional hint?
Btw, 0 < α < 1
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,291
If z7 = y, then z = y1/7
 

peterfede

New member
Joined
Oct 2, 2020
Messages
9
Thanks Jomo for your reply.
However this case is a bit more complicated.
There's the x even in the right side of the equation. If you read my last message, you see that I've already applied this basic rule.
Hoping someone can help me
 

Maurice

New member
Joined
Oct 14, 2020
Messages
5
Thanks Jomo for your reply.
However this case is a bit more complicated.
There's the x even in the right side of the equation. If you read my last message, you see that I've already applied this basic rule.
Hoping someone can help me
Hi, I have attached a file for you.
 

Attachments

peterfede

New member
Joined
Oct 2, 2020
Messages
9
Brilliant, I appreciate that. Thanks
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,291
Thanks Jomo for your reply.
However this case is a bit more complicated.
There's the x even in the right side of the equation. If you read my last message, you see that I've already applied this basic rule.
Hoping someone can help me
You were told to divide by x, so there is no x on the right hand side anymore. You were the one who asked for the next hint. I supplied one and you think that I am talking about the starting move.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,291
Hi, I have attached a file for you.
In the future please do not solve problems for students. We do not do that on this forum.
 

Maurice

New member
Joined
Oct 14, 2020
Messages
5
In the future please do not solve problems for students. We do not do that on this forum.
Sorry, I was not aware of that. But he seemed to have been struggling for a long time and to be unaware of the rules of indices. However, I will not do that again.
 
Top