Solving simultaneous equations

  • Thread starter chwala
  • Start date
  • #1
chwala
Gold Member
871
70
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
 

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
14
Okay, let's see. So just to make sure I have the problem clear... your two equations are [itex]2^{x+y} = 6^{y}[/itex] and [itex]3^{x} = 6 \cdot 2^{y}[/itex]. Fun looking problem, really. Well, let's see... if you multiply both sides of the second equation by 2^x, you get this:

[tex]6^{x} = 6 \cdot 2^{x+y}[/tex]

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

[tex]6^{x} = 6 \cdot 6^{y}[/tex]

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.
 
  • Like
Likes chwala
  • #3
chwala
Gold Member
871
70
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
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
961
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, [itex]a^x[/itex], has the inverse function, [itex]log_a(x)[/itex]. If [itex]a^x= a^y[/itex] then [itex]ln(a^x)= ln(a^y)[/itex] so that x= y.
 
  • #5
Char. Limit
Gold Member
1,204
14
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, [itex]a^x[/itex], has the inverse function, [itex]log_a(x)[/itex]. If [itex]a^x= a^y[/itex] then [itex]ln(a^x)= ln(a^y)[/itex] 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.
 

Related Threads on Solving simultaneous equations

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
21
Views
2K
Replies
1
Views
2K
Replies
9
Views
905
Replies
10
Views
9K
Replies
7
Views
3K
Replies
6
Views
821
Replies
2
Views
1K
Replies
7
Views
5K
  • Last Post
Replies
1
Views
2K
Top