I'm quite ashamed that I'm at a math-related

Angelina Kaufman

Angelina Kaufman

Answered question

2022-04-07

I'm quite ashamed that I'm at a math-related course at the university and I'm stuck. I can't solve at all this equation:
n=8log2(n).
I have tried applying the log property so it becomes 2n=28n
Besides that this didn't help me, I'm not even sure if I applied right the property that says x=log2(y)2x=y
Thanks in advance! Best Regards,

Answer & Explanation

tempur8x43

tempur8x43

Beginner2022-04-08Added 16 answers

1.No, you didn't apply it correctly. Since 8log2(n)=log2(n8) what you have is equivalent to n=log2(n8), or 2n=n8 Equivalently, by taking 8th roots, 2n8=n which can also be obtained by writing the original equation as n8=log2(n)
2.The reason you are having trouble is that there is no elementary solution. You can write a solution using Lambert's W function.
The W function has the property that W(z)=x if and only if z=xex.
To use Lambert's W function, we can proceed as follows: let u=n8 Then:
n=2n8
8u=2u
8u2u=1
1=8ueuln(2)
18=ueuln(2)
ln28=uln(2)euln(2)
W(ln28)=uln(2)
u=1ln(2)W(ln28)
n8=1ln(2)W(ln28)
n=8ln2W(ln28)
Added. Note that Lambert's W function on the reals is "double valued" on (1e,0). Since ln28 lies in this interval, this actually gives you two values for n.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?