Simple Quick Proof

[morvornagh]

if y=x^(n-1), what doe n equal

 

[srmichael]

if y=x^(n-1), what doe n equal

This makes no sense. What are you trying to do here?

 

[Deleted member 4993]

if y=x^(n-1), what doe n equal

Please read first:

http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting?p=322217#post322217

Then come back and ask your question again

 

[Bob Brown MSEE]

Hi morvornagh

if y=x^(n-1), what doe n equal

Welcome to FMH!
I'm not sure but I believe you are trying to solve for n in terms of x and y.

Hint: take log base x of both sides

 

[morvornagh]

Welcome to FMH!
I'm not sure but I believe you are trying to solve for n in terms of x and y.

Hint: take log base x of both sides

Thanks Bob - I am trying to solve for n in terms of x and y

so

if y=x^(n-1) then you are saying that ln y = ln (x^(n-1)) correct? But how then do I then get to what n-1 equals in terms of x and y.. I would like to get to an answer for n for a given x and y. Many Thanks

Apologies for not stating the original question more clearly i believe it should be fairly striaght forward but its been a while since I have had to solve for non-linear equations.

I am a little older and just need a little help thanks

 

[Bob Brown MSEE]

2. ֺ Tell us something about your age and math education.

Examples: "I'm in seventh grade - we're learning square roots"
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[lookagain]

Thanks Bob - > > > I am trying to solve for n < < <

. . . . . No, morvornagh, you still have not stated the problem correctly.
It is to solve for n in terms of x and y.

The prompt was stated to you in this post's fourth post.

so


Apologies for not stating the origginal question more clearly i believe it should be fairly striaghtforward
but its been a while since I have had to solve for non-linear equations.

...

 

[morvornagh]

...

Thanks

 

[Bob Brown MSEE]

ln(k)
---- = Log_x(k)
ln(x)

 

[morvornagh]

Non Linear Equation

I am trying to solve for n in terms of x and ywhere y=x^(n-1)

My question is how do I get to what n-1 equals in terms of x and y?... I would like to get to an answer for n for a given x and y. Many Thanks

I am a little older and just need a little help (please be gentle in your response)

thanks in advance

(I am re-posting this as I was kindly advised to provide more info on what I was trying to solve for.)

 

[Deleted member 4993]

I am trying to solve for n in terms of x and ywhere y=x^(n-1)

My question is how do I get to what n-1 equals in terms of x and y?... I would like to get to an answer for n for a given x and y. Many Thanks

I am a little older and just need a little help (please be gentle in your response)

thanks in advance

(I am re-posting this as I was kindly advised to provide more info on what I was trying to solve for.)

Hint:

Use

log(a^m) = m * log(a)

.

 

[JeffM]

Thanks Bob - I am trying to solve for n in terms of x and y

so

if y=x^(n-1) then you are saying that ln y = ln (x^(n-1)) correct? But how then do I then get to what n-1 equals in terms of x and y.. I would like to get to an answer for n for a given x and y. Many Thanks

Apologies for not stating the original question more clearly i believe it should be fairly striaght forward but its been a while since I have had to solve for non-linear equations.

I am a little older and just need a little help thanks

\(\displaystyle ln(y) = ln\left(x^{(n - 1)}\right) \implies \)

\(\displaystyle ln(y) = (n - 1) * ln(x) \implies\)

\(\displaystyle n -1 = \dfrac{ln(y)}{ln(x)} \implies \)

\(\displaystyle n = 1 + \dfrac{ln(y)}{ln(x)} \implies \)

\(\displaystyle n = \dfrac{ln(x) + ln(y)}{ln(x)} \implies\)

\(\displaystyle n = \dfrac{ln(xy)}{ln(x)}.\)

 

[morvornagh]

Hint:

Use

log(a^m) = m * log(a)

.

Perfect - Thank you so much

 

[mmm4444bot]

I am re-posting this as I was kindly advised to provide more info on what I was trying to solve for.

Please place additions to your conversation into the same thread. Do not duplicate exercises across multiple threads, or we will all need to navigate through a single discussion held at multiple locations, and that's both a recipe for disaster and a waste of time.

I've moved all of the posts related to this exercise into this thread.

Cheers :cool:

 

[morvornagh]

\(\displaystyle ln(y) = ln\left(x^{(n - 1)}\right) \implies \)

\(\displaystyle ln(y) = (n - 1) * ln(x) \implies\)

\(\displaystyle n -1 = \dfrac{ln(y)}{ln(x)} \implies \)

\(\displaystyle n = 1 + \dfrac{ln(y)}{ln(x)} \implies \)

\(\displaystyle n = \dfrac{ln(x) + ln(y)}{ln(x)} \implies\)

\(\displaystyle n = \dfrac{ln(xy)}{ln(x)}.\)

wow - thx so much

 

[morvornagh]

Please place additions to your conversation into the same thread. Do not duplicate exercises across multiple threads, or we will all need to navigate through a single discussion held at multiple locations, and that's both a recipe for disaster and a waste of time.

I've moved all of the posts related to this exercise into this thread.

Cheers :cool:

got it thanks