SOLUTION: Solve for all integer solutions of x and y. 3^x + 3^y = 108

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Question 1209372: Solve for all integer solutions of x and y.
3^x + 3^y = 108
Found 3 solutions by ikleyn, mccravyedwin, Edwin McCravy:
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.

    3%5E0 = 1,

    3%5E1 = 3,

    3%5E2 = 9,

    3%5E3 = 27,

    3%5E4 = 81,

    3%5E5 = 243.


At this point, it should be clear that there is no need to consider degrees higher than 3%5E4 = 81.


Trial and error method gives two solutions

    27 + 81 = 108,   81 + 27 = 108,

that are

    3%5E3 + 3%5E4 = 108,  3%5E4 + 3%5E3 = 108.


So, possible pairs (x,y) are  (3,4)  and  (4,3).    ANSWER

Solved.


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Commenting Edwin's post, the "trial and error" is the method which children learn intuitively
at the kindergarten age before they come to school and before they start learning Math.



Answer by mccravyedwin(408) About Me  (Show Source):
You can put this solution on YOUR website!
Trial and error may not be obvious to you, so you
might try doing a little algebra to make it easier
to see what to do.  



Divide both sides by 33 by subtracting
3 from both exponents of 3 on the left:



Factor out 3x-3 on the left:



That reminds us a little of 1%2A%281%2B3%29%22%22=%22%224.

That would be the case if the power of 3 before the parentheses 
were equal to 1, and the power of 3 inside the parentheses were
31.  

Now we know that 30 = 1.  And we see that x = 3 would make that
exponent = 0.

So now all we need to do is find y so that y-x = 1.  Substituting 
x = 3,  y - 3 = 1, or y = 4.

So  x = 3 and y = 4 are the answers.  Checking:



Edwin

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Commenting Ikleyn's post, equations involving variable exponents are not taught
in kindergarten.

Edwin