Asked by Thomas
Find x if 2^(x+1)+2^x=3^(y+2)-3^y and x and y are integers. Thanks in advance.
Answers
Answered by
Reiny
2^(x+1)+2^x=3^(y+2)-3^y
2^x(2^1 + 1) = 3^y(3^1 -1)
3 * 2^x = 8*3^y
2^x/3^y = 8/3
since x and y are to be integers, let's match up ...
2^x = 8 ----> x=3
and
3^y = 3 ----> y=1
check:
LS = 2^(x+1)+2^x = 2^4 + 2^3 = 24
RS = 3^(y+2)-3^y = 3^3 - 3^1 = 24
my answer is correct
2^x(2^1 + 1) = 3^y(3^1 -1)
3 * 2^x = 8*3^y
2^x/3^y = 8/3
since x and y are to be integers, let's match up ...
2^x = 8 ----> x=3
and
3^y = 3 ----> y=1
check:
LS = 2^(x+1)+2^x = 2^4 + 2^3 = 24
RS = 3^(y+2)-3^y = 3^3 - 3^1 = 24
my answer is correct
Answered by
Reiny
typo in line 2, does not affect the rest of the solution
should be:
2^x(2^1 + 1) = 3^y(3^2 -1)
should be:
2^x(2^1 + 1) = 3^y(3^2 -1)
Answered by
oobleck
when we got to this step:
3 * 2^x = 8*3^y
I wasn't motivated to proceed as Reiny did.
I'd have proceeded
2^x * 3^1 = 2^3 * 3^y
By the Fundamental Theorem of Arithmetic, a number has a unique prime factorization. So, the exponents of 2 and 3 must match up.
Maybe this was implied in Reiny's fraction, which was reduced to lowest terms...
3 * 2^x = 8*3^y
I wasn't motivated to proceed as Reiny did.
I'd have proceeded
2^x * 3^1 = 2^3 * 3^y
By the Fundamental Theorem of Arithmetic, a number has a unique prime factorization. So, the exponents of 2 and 3 must match up.
Maybe this was implied in Reiny's fraction, which was reduced to lowest terms...
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