# Thread: Solve 10 log_2 (n) = n^(1/2)

1. ## Solve 10 log_2 (n) = n^(1/2)

Hi,

Please let me know how to solve the below equation

10 log n = n^(1/2)

Base : 2
n^ 1/2 is n raise to the power 1/2.

It can be solved by substituting values but I need a mathematical solution.
Thanks

2. You cannot solve it algebraically - your best bet is to solve via inspection which is what you've done.

Sketching the graphs on the same axes shows that there is only one solution since $10 \log_2 (n)$ grows faster than $\sqrt{n}$

It would appear the solution is quite close to $(1.0745,1.0366)$

3. Originally Posted by athap
Hi,

Please let me know how to solve the below equation

10 log n = n^(1/2)

Base : 2
n^ 1/2 is n raise to the power 1/2.

It can be solved by substituting values but I need a mathematical solution.
Thanks
Where has this equation come from? It helps if all of the question is given (including perhaps whether an approximate solution is sufficient). An exact solution can probably be found using the Lambert W-function (a function I doubt you have ever heard of).

4. Thanks GC and Mr Fantastic.

An approximate answer would be ok otherwise the equation would become quite complex.

5. Mr Fantastic you are right.

Can u please tell me how to proceed with W(-ln 2/10) .i.e the Lambert W function

6. Originally Posted by athap
Mr Fantastic you are right.

Can u please tell me how to proceed with W(-ln 2/10) .i.e the Lambert W function
What have you tried?

7. I transformed the equation to Y = X e^X <==> X = W(Y) form.

Am I making any sense??

Source of this is Wikipedia

8. Originally Posted by athap
I transformed the equation to Y = X e^X <==> X = W(Y) form.

Am I making any sense??

Source of this is Wikipedia
I don't think Lambert's W is covered by any Pre-Algebra course in the multi-verse, so I would suggest you do the other thing that MrF suggested and post the actual question and/or its' context.

CB