Solving simultaneous equations

[chwala]

I am trying to solve the simultaneous equation 2^x+y = 6^y and 3^x = 6(2^y) i have solved as follows
3^x . 2^x+y = 3^x . 6^y
3^x. 2^x+y = 6(2^y). 2^x+y on subtraction
i get
3^x.6^y - 6(2^y).2^x+y = 0 then
3^x. 6^y = 6(2^y). 2^x. 2^y on solving i get
x log (3/2)= log 6+y log(3/2) now i do i get x and y from this step. the answers are x=2.71 and y = 1.71 regards

 

[Char. Limit]

Okay, let's see. So just to make sure I have the problem clear... your two equations are $2^{x+y} = 6^{y}$ and $3^{x} = 6 \cdot 2^{y}$. Fun looking problem, really. Well, let's see... if you multiply both sides of the second equation by 2^x, you get this:

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

Then if we substitute the second equation in for the first, we get this:

$$6^{x} = 6 \cdot 6^{y}$$

Now this looks nice and simple. The right side simplifies to 6^(y+1), and so we have 6^(x) = 6^(y+1). Now, there's a certain property, I can't remember what it is, but basically it says if a^x = a^y, then x=y. So using that property, we have x=y+1. Good! We found a simple linear relation between x and y. Now we just need to figure out what satisfies this. To solve that, we substitute y+1 wherever we see an x in the first equation to find the solution for x, which we then need to subtract 1 from that solution to get y. That should be the whole solution set.

 

[chwala]

Thanks i now get it
2^x+y =6^y since x=y+1,
2^y+1+y = 6^y
2^2y+1 = 6^y
4^y+2 = 6^y
log 2 = y log (6/4)
y= log 2/ log 1.5
y= 1.709 hence x = 1.709+1 = 2.709

 

[HallsofIvy]

Now, there's a certain property, I can't remember what it is, but basically it says if a^x = a^y, then x=y. So using that property, we have x=y+1.

The property is that the function is "one-to-one": if f is a "one-to-one" function and f(x)= f(y) then x= y. Or, equivalently, that the exponential function, $a^x$, has the inverse function, $log_a(x)$. If $a^x= a^y$ then $ln(a^x)= ln(a^y)$ so that x= y.

 

[Char. Limit]

The property is that the function is "one-to-one": if f is a "one-to-one" function and f(x)= f(y) then x= y. Or, equivalently, that the exponential function, $a^x$, has the inverse function, $log_a(x)$. If $a^x= a^y$ then $ln(a^x)= ln(a^y)$ so that x= y.

Ah, thank you, that's the one! Of course, the exponential function is only one-to-one if we assume x and y are both real, but we can safely assume that here.

 

[chwala]

I looked at this question again, though tedious we can also have this approach,

$6 (2^y)⋅ 2^{x+y} = 6^y⋅ 6 (2^y)$
$6^y ⋅3^x = 6 (2^y)⋅6^y$

then,
$6 (2^y)⋅ 2^{x+y} = 6^y ⋅3^x$

$\dfrac{2^y ⋅ 2^x}{3^x ⋅ 3^y} = \dfrac {1}{6}$

$\left(\dfrac{2}{3}\right)^{x+y} = \dfrac {1}{6}$

$(x+y) \log \dfrac{2}{3} = \log \dfrac {1}{6}$

$(x+y)=\dfrac{\log \dfrac{1}{6}}{\log \dfrac{2}{3}}$

$x+y = 4.419$ to three decimal places.

from,

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

$(x+y) \log 2 = y \log 6$

$4.419 \log 2 = y \log 6$

$y = \dfrac {4.419 \log 2}{\log 6}$

$y = 1.71$

$x=4.419-1.71= 2.71$ to two decimal places.