Solve the simultaneous equations

[chwala]

Homework Statement

$2^{x+y}=6^y$

$3^x=6(2^y)$

Relevant Equations

simultaneous equations

$2^{x+y+1}$=$3^{x+y-1}$
$\frac {x+y+1}{x+y-1}$=$\frac {log 3}{log 2}$
this is where i reached, i just got this problem from my old notes, i do not think we can solve this...

 

[Mark44]

Homework Statement:: $2^{x+y}=6^y$

$3^x=6.2^y$
Relevant Equations:: simultaneous equations

$2^{x+y+1}$=$3^{x+y-1}$
$\frac {x+y+1}{x+y-1}$=$\frac {log 3}{log 2}$
this is where i reached, i just got this problem from my old notes, i do not think we can solve this...

Take the log (in whatever is your favorite log base) of both sides of each equation.

 

[chwala]

alternatively,
$2^\frac {x+y}{y}$=$6$
$2^\frac {x+y}{y}$=$\frac{3^x}{2^y}$
$2^{x+y}=3^x$
$(x+y) log 2$= $xlog 3$
on simplifying this i end up with,
$0.5849x=y$

 

[chwala]

Take the log (in whatever is your favorite log base) of both sides of each equation.

ok, but will still need to express $x$ in terms of $y$...right, just check my subsequent working on this...

 

[Mark44]

Once you have solved for x in terms of y or y in terms of x, just substitute in the other equation.

 

[chwala]

ok, i get
$3^x=6.2^y$
$xlog 3= log 6+0.5849x log 2$
$0.477x=0.778+0.176x$
$0.477x-0.176x=0.778$
$0.301x=0.778$
$x=2.58$
is this correct? actually i am trying to prove the values i.e
$x=2.5862$
$y=1.5142$
and i can see that these values do not approximate the first equation of the problem...
but interestingly, they satisfy the equation,
$2^{x+y}=3^x$ which follows from the original problem...

 

[Dr Transport]

$$x\log(3) = y\log(6.2)$$
$$(x+y)\log(2)= x\log(3)$$

now solve for y in the first equation and substitute into the second and solve for x.

 

[chwala]

I think I already did that...check my posts...

 

[chwala]

What i meant in post $6$, is that the values of $x$ and $y$ do not satisfy the original equation, $2^{x+y}$=$6^y$, but satisfy the other equation $2^{x+y}=3^x$, that follows from the original problem. I hope I am clear...

 

[Mark44]

ok, i get
$3^x=6.2^y$
$xlog 3= log 6+0.5849x log 2$

I can't tell what you did in the equation above.
If $3^x = 6.2^y$, then $x \ln(3) = y \ln(6.2)$

$0.477x=0.778+0.176x$
$0.477x-0.176x=0.778$
$0.301x=0.778$
$x=2.58$
is this correct? actually i am trying to prove the values i.e
$x=2.5862$
$y=1.5142$

Where did these values come from?
If they don't satisfy both equations, they are not a solution to the system of equations.

and i can see that these values do not approximate the first equation of the problem...
but interestingly, they satisfy the equation,
$2^{x+y}=3^x$ which follows from the original problem...

It's really irrelevant that they happen to satisfy only one of the equations. Since it is an equation in two unknowns, there are an infinite number of pairs of numbers that satisfy it.

For the record, I get y = 0 and x = 0, which satisfy both equations.

 

[chwala]

Sorry Mark, that equation is supposed to be $3^x$=$6$times $2^y$ and not $6.2^y$

 

[chwala]

allow me to amend the original equation...