given u = 2vx^2, how can i use ln and turn it into linear equation? so far i have this?

Question

I need to get the slope to be somehow related to v. the slope should either be v or ln v or log v or something. remember i need this to be in the form y=mx+b.

so far...
ln u = ln 2 + 2 (ln vx).

basically i should be able to calculate v using the slope.

cattbarf i know what you are saying but v should not be part of the x but part of the m in y = mx + b.

Answer 1

ln u = ln (2vx²)
ln u = ln (2v) + ln (x²)
ln u = ln (2v) + 2 ⋅ ln (x)

If you allow for a little substitution:
y = 2⋅x + ln (2v)
Where x in this equation isnt the same as x in the original.
y = ln u and x = ln x (left half of these equations is for the new equation, right is from the old)

You cannot solve for any of these variables because you only have one equation containing three variables.

Answer 2

so far you have the quantity it is going to value to construct the turbine and perform it for x days. All thats lacking is the revenues area of the equation: in case you may sell the potential for $500.00 in line with day, then revenues is 500x. The question is how many days it is going to take to get returned the funds they invested, so we set y equivalent to the value minus the revenues (or the income, in different words) y = 150x + 20000 - 500x y = -350x + 20000 The form of days till at last the turbine will pay for itself stands out as the value of x while y = 0. (in actuality, interior the genuine international, you will possibly additionally incorporate a internet present value calculation to this, which takes under consideration the quantity you will possibly have made by potential of leaving your shape and working expenditures in an pastime bearing account, yet shall we not complicate this...)

Answer 3

If ln u = Y and ln vx = X, and ln2 = 0.693, what does Y= 0.693 + 2X remind you of?