Solving simultaneous equations

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The discussion focuses on solving the simultaneous equations 2^(x+y) = 6^y and 3^x = 6(2^y). A key step involves recognizing that if 6^x = 6^(y+1), then x must equal y + 1. Substituting this relationship into the first equation simplifies the problem, leading to the conclusion that y can be expressed as log(2)/log(1.5). Ultimately, the solutions are found to be approximately x = 2.71 and y = 1.71. The conversation highlights the use of logarithmic properties to derive these results effectively.
chwala
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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
 
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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.
 
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
 
Char. Limit said:
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.
 
HallsofIvy said:
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.
 
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.
 
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