Find x if 2^(x+1)+2^x=3^(y+2)-3^y and x and y are integers. Thanks in advance.
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
typo in line 2, does not affect the rest of the solution
should be:
2^x(2^1 + 1) = 3^y(3^2 -1)
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...
To find the value of x, we need to solve the equation: 2^(x+1) + 2^x = 3^(y+2) - 3^y
Let's simplify the equation step by step:
First, let's simplify the left side of the equation using the rules of exponents:
2^(x+1) + 2^x = 2 * 2^x + 2^x = 2^x(2 + 1) = 3 * 2^x
Now, let's simplify the right side of the equation using the rules of exponents:
3^(y+2) - 3^y = 3^y * 3^2 - 3^y = 3^y(9 - 1) = 8 * 3^y
Now our equation becomes: 3 * 2^x = 8 * 3^y
To solve this equation, we can rewrite 3 and 8 as powers of 2:
3 * 2^x = 8 * 3^y
3 * 2^x = 2^3 * 3^y
2^x * 3 = 2^3 * 3^y
Comparing the exponents of 2 on both sides, we can set up an equation:
x = 3
Now, let's substitute this value back into the original equation to find the value of y:
2^(x+1) + 2^x = 3^(y+2) - 3^y
2^(3+1) + 2^3 = 3^(y+2) - 3^y
2^4 + 2^3 = 3^(y+2) - 3^y
16 + 8 = 3^(y+2) - 3^y
24 = 3^(y+2) - 3^y
Now, we need to solve this equation for y.