# can´t solve it this simple equation

• Apr 19th 2008, 02:48 AM
gruvan
can´t solve it this simple equation
Hello
I have a equation with y as dependent variable but want instead x as the dependent variable. I can´t solve it (Headbang) can you?

y=(((10^2.806*x^-0.444*11.08^0.138*4202463^0.235*5508000^0.4)*1000)/0.8515)/1000000

Thanks guys for all the help! but if I simplify the equation to just
y=10^2.806*x^-0.444*b^0.138*c^0.235*d^0.4
what would the equation then be with x as the dependent?

x=................................................ .................................................. ..
• Apr 19th 2008, 03:18 AM
topsquark
Quote:

Originally Posted by gruvan
Hello
I have a equation with y as dependent variable but want instead x as the dependent variable. I can´t solve it (Headbang) can you?

y=(((10^2.806*x^-0.444*11.08^0.138*4202463^0.235*5508000^0.4)*1000)/0.8515)/1000000

x=................................................ .................................................. ..

According to what you have written, this is
$y = \frac{1000 \cdot 10^{2.806x^{-0.444 \cdot 11.08^{0.138 \cdot 4202463^{0.235 \cdot 5508000^{0.4}}}}}}{0.8515 \cdot 1000000}$

There is a way to solve for x but as far any practical calculation is concerned,
$z = 11.08^{0.138 \cdot 4202463^{0.235 \cdot 5508000^{0.4}}}$
is so large that
$x^{-0.444 \cdot z} \approx 1$
to some 1000 decimal places.

So effectively y is a constant function for any reasonable scale factor.

Edit: I suppose for completeness I should say that this equation is basically just
$y = a \left ( 10^{x^b} \right )$
so the solution would be
$\frac{y}{a} = 10^{x^b}$

$x^b = log \left ( \frac{y}{a} \right )$

$x = \left [ log \left ( \frac{y}{a} \right ) \right ] ^{1/b}$

-Dan
• Apr 19th 2008, 06:28 AM
CaptainBlack
Quote:

Originally Posted by topsquark
According to what you have written, this is
$y = \frac{1000 \cdot 10^{2.806x^{-0.444 \cdot 11.08^{0.138 \cdot 4202463^{0.235 \cdot 5508000^{0.4}}}}}}{0.8515 \cdot 1000000}$

There is a way to solve for x but as far any practical calculation is concerned,
$z = 11.08^{0.138 \cdot 4202463^{0.235 \cdot 5508000^{0.4}}}$
is so large that
$x^{-0.444 \cdot z} \approx 1$
to some 1000 decimal places.

So effectively y is a constant function for any reasonable scale factor.

Edit: I suppose for completeness I should say that this equation is basically just
$y = a \left ( 10^{x^b} \right )$
so the solution would be
$\frac{y}{a} = 10^{x^b}$

$x^b = log \left ( \frac{y}{a} \right )$

$x = \left [ log \left ( \frac{y}{a} \right ) \right ] ^{1/b}$

-Dan

By the usual rules of precedence ^ take precedence over * so this would be
(under this interpretation, which I would not trust. It really does need brackets when written in plain ASCII):

$
y=(10^{2.806} x^{-0.444} 11.08^{0.138} 4202463^{0.235} 5508000^{0.4}1000)/(0.8515 \times 1000000)$

Either way the original poster needs to learn how cancell off some of the zeros and break the calculation down into manageble chunks and not to try to calculate the value of this monster as it is (else they are just taking the p**s)

for all the good it will do:

x= 2.38221e-010 y^{-2.25225}

RonL
• Apr 20th 2008, 12:38 PM
gruvan
Thanks guys for all the help! but if I simplify the equation to just
y=10^2.806*x^-0.444*b^0.138*c^0.235*d^0.4
what would the equation then be with x as the dependent?
x=................................................ ...........................
• Apr 20th 2008, 06:40 PM
topsquark
Quote:

Originally Posted by gruvan
Thanks guys for all the help! but if I simplify the equation to just
y=10^2.806*x^-0.444*b^0.138*c^0.235*d^0.4
what would the equation then be with x as the dependent?
x=................................................ ...........................

Well everything but the x term is just a constant. So your equation is of the form
$y = Ax^b$

$x = \left ( \frac{y}{A} \right ) ^{1/b}$

-Dan