Asked by Akbar

Can somebody solve this system of euation
:2^x+3^y=17 2^x+2+3^y+1=5
Answered by Steve
without trying complicated algebraic solutions, you know that

17 = 8+9 = 2^3 + 3^2

Now just check to make sure that also works in the other equation.

It does not: 2^3 + 2 + 3^2 + 1 is not 5.

If you mean

2^(x+2) + 3^(y+1) = 5

then since 2^1 + 3^1 = 5,
x = -1 and y = 0

I can't think of any numbers that satisfy both equations.
Answered by Reiny
Surely for the 2nd equation Akbar must have meant:
2^(x+2)+3^(y+1)=5
or else why not just add up those constants.

(2^2)(2^x) + 3(3^y) = 5

let a = 2^x and let b = 3^y, then we can reduce our equations to ...

a+b = 17
4a + 3b = 5

first one by 3
3a + 3b = 51
4a + 3b = 5
a = -46
b = 63

then 2^x = -46 , which is not possible in the real number set

So for 2^x + 3^y = 17 to be true, we would have add a "non-existing number" to 3^y to get 17.

So agreeing with Steve, there is no real solution.
http://www.wolframalpha.com/input/?i=plot+2%5Ex%2B3%5Ey%3D17+%2C+2%5E%28x%2B2%29%2B3%5E%28y%2B1%29%3D5
Answered by sushant
2^3+3^2=17
2^(3+2)-3^(2+1)=5
Answered by Einstein
x=3 and y=2
Answered by Nameer
Let 2^x be A
3^y be B
Then A+B=17 ----eq1
And 2^x X 2^2 — 3^y X 3=5
A X 4 — B X 3=5
4A—3B=5 ----eq2
Solve eq1 and eq2
A=8 and B=9
Then 2^x=8 and 3^y=9
2^x=2^3 3^y=3^2

Therefore x=3 and y=2
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