math genious anyone?

I have a formula I am trying to solve for a different variable in it.

x = ( l + 300 * 2^l/7) / 4

I want to get L by itself, so i divide x off the one side

0 = ((l + 300 * 2^l/7) / 4) * x

Then divide l from the right side

1/l = ((300*2^l/7) / 4)) * x

now is the part i dont understand i have to do the oposite of 2^l/7
which is a radical. Little ( 7 Squareroot 2) ^ l

(1/l) (7 squareroot 2) ^l = (300 / 4) * x

how do I add the right side together or am I doing this totally wrong??

 

i thought what you do to one side you have to do to the other...

anyways, heres what i got

x= ((L+300*2^(L/7)/4)*x
multiply through by 4
4x = (l+300*2^(L/7)*4x
divide by 4x
1=L+300*2^(L/7)

2^(L/7) = (7th root of 2)^L
so
1=L+300*(365.7040963)^L
i put this into my calculator :-D
L= 7.85142857

byu

 

Neither of these responses are correct! Dividing or multiplying both sides of x = ( l + 300 * 2^l/7) / 4 (if this is indeed the correct equation to solve) by x (or l for that matter) does not yield to any of the initial equations shown and will not lead to a solution of l in terms of x. Furthermore, nonsense solutions will occur as x can be zero and you can end up proving 2=1, e.g., consider when l = 75 then x = 0.

Assume 2^l/7 is 2 to the power of l/7, i.e., 2^(l/7) rather than 2 to the power of l all divided by 7, i.e., (2^l)/7 - not that it makes any difference to solving.

Also assume, that it is l + rather than 1 +?


The solution goes:

x = ( l + 300 * 2^(l/7)) / 4

Multiply both sides by 4:

4x = l + 300 * 2^(l/7)

Then it becomes non-trivial and you probably need to take logarithms base 2 on each side to eliminate 2^, as this is the inverse function. Note that logarithms require positive real numbers, unless you're going into complex numbers. This would than make assumptions about x and l that we don't know about regarding their numerical ranges.

What makes it non-trivial is the first term of l + as the logarithm encloses this.

Note that log (a^p) = p log(a) and log (a b) = log (a) + log (b).

 

Not sure how that stuff works. I made a mistake though, I dont want to solve for L I just want the equation to be L = whatever it may be. I tried this in a equation calculator and I got those log variables and I dont understand what they are, and in the end. The equation is: x = ( l + 300 * 2^(l/7)) / 4

 

Solving for L in terms of X is the same as obtaining an equation as L =. That is, if you don't mind L on both sides of the equation.

For example:

4x = l + 300 * 2^(l/7)

Then subtracting l from both sides gives:

4x - l = 300 * 2^(l/7)

Then subtracting 4x and multiplying by -1 from sides gives:

l = 4x - 300 * 2^(l/7)

Hence this satisfies your request to obtain and equation as L = but I sense that this is not what you are after.

I believe that you want an equation for L in terms of X and thus need to eliminate all L's on the right hand side to end up with L = f(X) whatever functoin f(x) is. It is clear from the graph that f(X) is logarithmic and I'm guessing that it is something like:

a log(x - b) - c

where a, b and c are constants. That is:

l = a log(x - b) - c

 

AHHH! Thanks bud, I figured it out finally, you were right

 

What's a 'genious'?:confused - Must be a generous genius. LOL