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Convert funktion to variable

Posted 9 years ago

Hi, I don't get any results by using the therm "solve 1=(3/4)((F+k(l-a))/(ba^(1.5)*r^(0.5))) for a" in wolframalpha. The window with the results is empty and really need this solution. What I'm doing wrong?

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POSTED BY: Stefan K.
4 Replies
Posted 9 years ago

Ok now I got it. The problem were the spaces instead of multiplication sign ;)

POSTED BY: Stefan K.
Posted 9 years ago

Wow thank you :)

I tried the same (Solve[(3/4)((F+k(l-a))/(ba^(1.5)*r^(0.5))) == 1, a]) in Chrome, Edge and Firefox but every time I got an empty result. For other solutions, I get them. What could be the reason to get an empty page for this issue?

POSTED BY: Stefan K.

That works

enter image description here

but says that standard computation time exceeded. There are three solutions for a.

POSTED BY: Udo Krause

Mathematica says

In[49]:= Remove[F, k, l, a, b, r]
Solve[(3/4) ((F + k (l - a))/(b a^(3/2) r^(1/2))) == 1, a]

Out[50]= {{a -> (3 k^2)/(
    16 b^2 r) - (-((81 k^4)/(b^4 r^2)) + (864 k (F + k l))/(
       b^2 r))/(144 3^(
       1/3) ((9 k^6)/(b^6 r^3) - (144 F k^3)/(b^4 r^2) - (144 k^4 l)/(
         b^4 r^2) + (384 F^2)/(b^2 r) + (768 F k l)/(b^2 r) + (
         384 k^2 l^2)/(b^2 r) + (
         64 Sqrt[3]
           Sqrt[-F^3 k^3 - 3 F^2 k^4 l - 3 F k^5 l^2 - k^6 l^3 + 
           12 b^2 F^4 r + 48 b^2 F^3 k l r + 72 b^2 F^2 k^2 l^2 r + 
           48 b^2 F k^3 l^3 r + 12 b^2 k^4 l^4 r])/(b^3 r^(3/2)))^(
       1/3)) + 1/16 3^(
     1/3) ((9 k^6)/(b^6 r^3) - (144 F k^3)/(b^4 r^2) - (144 k^4 l)/(
       b^4 r^2) + (384 F^2)/(b^2 r) + (768 F k l)/(b^2 r) + (
       384 k^2 l^2)/(b^2 r) + (
       64 Sqrt[3]
         Sqrt[-F^3 k^3 - 3 F^2 k^4 l - 3 F k^5 l^2 - k^6 l^3 + 
         12 b^2 F^4 r + 48 b^2 F^3 k l r + 72 b^2 F^2 k^2 l^2 r + 
         48 b^2 F k^3 l^3 r + 12 b^2 k^4 l^4 r])/(b^3 r^(3/2)))^(
     1/3)}, {a -> (3 k^2)/(
    16 b^2 r) + ((1 + I Sqrt[3]) (-((81 k^4)/(b^4 r^2)) + (
         864 k (F + k l))/(b^2 r)))/(288 3^(
       1/3) ((9 k^6)/(b^6 r^3) - (144 F k^3)/(b^4 r^2) - (144 k^4 l)/(
         b^4 r^2) + (384 F^2)/(b^2 r) + (768 F k l)/(b^2 r) + (
         384 k^2 l^2)/(b^2 r) + (
         64 Sqrt[3]
           Sqrt[-F^3 k^3 - 3 F^2 k^4 l - 3 F k^5 l^2 - k^6 l^3 + 
           12 b^2 F^4 r + 48 b^2 F^3 k l r + 72 b^2 F^2 k^2 l^2 r + 
           48 b^2 F k^3 l^3 r + 12 b^2 k^4 l^4 r])/(b^3 r^(3/2)))^(
       1/3)) - 1/32 3^(
     1/3) (1 - I Sqrt[3]) ((9 k^6)/(b^6 r^3) - (144 F k^3)/(
       b^4 r^2) - (144 k^4 l)/(b^4 r^2) + (384 F^2)/(b^2 r) + (
       768 F k l)/(b^2 r) + (384 k^2 l^2)/(b^2 r) + (
       64 Sqrt[3]
         Sqrt[-F^3 k^3 - 3 F^2 k^4 l - 3 F k^5 l^2 - k^6 l^3 + 
         12 b^2 F^4 r + 48 b^2 F^3 k l r + 72 b^2 F^2 k^2 l^2 r + 
         48 b^2 F k^3 l^3 r + 12 b^2 k^4 l^4 r])/(b^3 r^(3/2)))^(
     1/3)}, {a -> (3 k^2)/(
    16 b^2 r) + ((1 - I Sqrt[3]) (-((81 k^4)/(b^4 r^2)) + (
         864 k (F + k l))/(b^2 r)))/(288 3^(
       1/3) ((9 k^6)/(b^6 r^3) - (144 F k^3)/(b^4 r^2) - (144 k^4 l)/(
         b^4 r^2) + (384 F^2)/(b^2 r) + (768 F k l)/(b^2 r) + (
         384 k^2 l^2)/(b^2 r) + (
         64 Sqrt[3]
           Sqrt[-F^3 k^3 - 3 F^2 k^4 l - 3 F k^5 l^2 - k^6 l^3 + 
           12 b^2 F^4 r + 48 b^2 F^3 k l r + 72 b^2 F^2 k^2 l^2 r + 
           48 b^2 F k^3 l^3 r + 12 b^2 k^4 l^4 r])/(b^3 r^(3/2)))^(
       1/3)) - 1/32 3^(
     1/3) (1 + I Sqrt[3]) ((9 k^6)/(b^6 r^3) - (144 F k^3)/(
       b^4 r^2) - (144 k^4 l)/(b^4 r^2) + (384 F^2)/(b^2 r) + (
       768 F k l)/(b^2 r) + (384 k^2 l^2)/(b^2 r) + (
       64 Sqrt[3]
         Sqrt[-F^3 k^3 - 3 F^2 k^4 l - 3 F k^5 l^2 - k^6 l^3 + 
         12 b^2 F^4 r + 48 b^2 F^3 k l r + 72 b^2 F^2 k^2 l^2 r + 
         48 b^2 F k^3 l^3 r + 12 b^2 k^4 l^4 r])/(b^3 r^(3/2)))^(
     1/3)}}
POSTED BY: Udo Krause
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