2^(n/8)=n

Solution for 2^(n/8)=n equation:

2^(n/8)=n

We move all terms to the left:

2^(n/8)-(n)=0

We add all the numbers together, and all the variables

2^(+n/8)-n=0

We add all the numbers together, and all the variables

-1n+2^(+n/8)=0

We multiply all the terms by the denominator


-1n*8)+2^(+n=0

We add all the numbers together, and all the variables

n-1n*8)+2^(=0

Wy multiply elements

-8n^2+n=0

a = -8; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-8)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-8}=\frac{-2}{-16}
=1/8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-8}=\frac{0}{-16}
=0
$