Logarithm Trouble With Two Variable Equation

Entropy

New member
Joined
Sep 2, 2012
Messages
5
Hello -- I am having trouble solving the following problem and would greatly appreciate some help.

Solve for n in terms of t:

10t = [ (3(n-5) * e(1.5n+5-3.2n)) / 2(n-3) ] * 10(2-n)

I have tried to solve this equation in Matlab and Wolfram Alpha, but can't seem to get the syntax right. If anyone could point me in the right direction I would be very grateful.
Many thanks,
Chris
 
Hello -- I am having trouble solving the following problem and would greatly appreciate some help.

Solve for n in terms of t:

10t = [ (3(n-5) * e(1.5n+5-3.2n)) / 2(n-3) ] * 10(2-n)

I have tried to solve this equation in Matlab and Wolfram Alpha, but can't seem to get the syntax right. If anyone could point me in the right direction I would be very grateful.
Many thanks,
Chris

You should not need Wolfram or Matlab for that.

Take log10 on both side - what do you get?

that is: solve

Log10[10t] = Log10[{ (3(n-5) * e(1.5n+5-3.2n)) / 2(n-3) } * 10(2-n)]

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
Thanks for the help. I would gladly share my work, however I am unsure of how to begin solving this equation.

Log10[10t] = Log10[{ (3(n-5) * e(1.5n+5-3.2n)) / 2(n-3) } * 10(2-n)]

It's been a long time since I have used logarithms. Any advice on the next step would be greatly appreciated.
 
Need to know the basic properties of logs, listed here.
http://www.freemathhelp.com/forum/threads/76893-Logarithim-Help

Thanks -- that was a very helpful link.

I was able to use the post at the bottom of the above link to get to the following:

t * Log10 (10) = Log10 (3n-5 * e5-1.7n) - Log10 (2n-3) + Log10 (102-n)

which reduces to:

t = Log10 (3n-5 * e5-1.7n) - Log10 (2n-3) + Log10 (102-n)

I am fairly confident about the right-hand side of the equal sign, but am unsure if I correctly evaluated the left-hand side. As well, I am unsure where to go next...
Thanks for all the help!
 
Thanks -- that was a very helpful link.

I was able to use the post at the bottom of the above link to get to the following:

t * Log10 (10) = Log10 (3n-5 * e5-1.7n) - Log10 (2n-3) + Log10 (102-n)

which reduces to:

t = Log10 (3n-5 * e5-1.7n) - Log10 (2n-3) + Log10 (102-n)

I am fairly confident about the right-hand side of the equal sign, but am unsure if I correctly evaluated the left-hand side. As well, I am unsure where to go next...
Thanks for all the help!

Your LHS is correct. However, RHS needs further expansion.

Now you need to use following rules of logarithm (which were stated in the referenced thread):

Loga(x*y) = Loga(x) + Loga(y)

Loga(xy) = y * Loga(x) and

Loga(a) = 1

These and some other properties are derived from the following definition of log:

If Loga(b) = c then ac = b while b > 0

and

[txt]a > 0 and a \ne 1[/txt]

thus

Log10 (3n-5 * e5-1.7n) = Log10 (3n-5) + Log10 (e5-1.7n)
 
Last edited by a moderator:
Thank you so very much!

My intuition leads me to believe that we should be using the natural logarithm, rather than log base 10. In that vein, would you mind letting me know if this looks right?

Starting point:

t ln(10) = ln(3n-5) + ln(e5-1.7n) - ln(2n-3) + ln(102-n)

Use log rules:

t ln(10) = (n-5)ln(3) + (5-1.7n)ln(e) - (n-3)ln(2) + (2-n)ln(10)

t = n ln(3) - 5ln(3) + 5ln(e) - 1.7n ln(e) + 2-n + n ln(2) -3ln(2)

Get all the n-terms on one side:

n ln(3) - 1.7ln(e) - n - n ln(2) = t + 5ln(3) - 5ln(e) + 3ln(2) - 2

Factor:

n [ln(3) - 1.7 - 1 + ln(2)] = t + 5ln(3) - 5ln(e) + 3ln(2) - 2

Solve for n:

n = [t + 5ln(3) - 5ln(e) + 3ln(2) - 7] / [ln(3) - 1.7 - 1 + ln(2)]

n = [t + 5ln(3) + 3ln(2) - 7] / [ln(3) + ln(2) - 2.7]

Is it reasonable to reduce this further?

n = (t + 1) / -2.7

Thanks again for all your help -- you guys have been a life-saver!
 
Thank you so very much!

My intuition leads me to believe that we should be using the natural logarithm, rather than log base 10. In that vein, would you mind letting me know if this looks right?

Starting point:

t ln(10) = ln(3n-5) + ln(e5-1.7n) - ln(2n-3) + ln(102-n)

Use log rules:

t ln(10) = (n-5)ln(3) + (5-1.7n)ln(e) - (n-3)ln(2) + (2-n)ln(10)

t = n ln(3) - 5ln(3) + 5ln(e) - 1.7n ln(e) + 2-n + n ln(2) -3ln(2) ----> What happened to ln(10) in the LHS. You have some sign problems in the RHS .... otherwise good.

Get all the n-terms on one side:

n ln(3) - 1.7ln(e) - n - n ln(2) = t + 5ln(3) - 5ln(e) + 3ln(2) - 2

Factor:

n [ln(3) - 1.7 - 1 + ln(2)] = t + 5ln(3) - 5ln(e) + 3ln(2) - 2

Solve for n:

n = [t + 5ln(3) - 5ln(e) + 3ln(2) - 7] / [ln(3) - 1.7 - 1 + ln(2)]

n = [t + 5ln(3) + 3ln(2) - 7] / [ln(3) + ln(2) - 2.7]

Is it reasonable to reduce this further?

n = (t + 1) / -2.7

Thanks again for all your help -- you guys have been a life-saver!

.
 
Thank you! In regard to your question, I divided both sides by ln(10) to eliminate it from the LHS and RHS... Is that not possible? Did I do it incorrectly?
Many thanks,
Entropy

Use log rules:


t ln(10) = (n-5)ln(3) + (5-1.7n)ln(e) - (n-3)ln(2) + (2-n)ln(10)

t = n ln(3) - 5ln(3) + 5ln(e) - 1.7n ln(e) + 2-n + n ln(2) -3ln(2) ----> What happened to ln(10) in the LHS. You have some sign problems in the RHS .... otherwise good.
 
Thank you! In regard to your question, I divided both sides by ln(10) to eliminate it from the LHS and RHS... Is that not possible ... Yes ? Did I do it incorrectly... Yes?
Many thanks,
Entropy

Use log rules:


t ln(10) = (n-5)ln(3) + (5-1.7n)ln(e) - (n-3)ln(2) + (2-n)ln(10)

t = n ln(3)/ln(10) - 5ln(3)
/ln(10) + 5ln(e)/ln(10) - 1.7n ln(e)/ln(10) + 2 - n - n ln(2)/ln(10) + 3ln(2)//ln(10) ----> good.

.
 
Top