MQ66 --- Exponential Equation

9ˣ + 3ˣ = 4ˣ + 2ˣ
==> (3ˣ)² + 3ˣ = (2ˣ)² + 2ˣ
==> (3ˣ)² + 3ˣ - (2ˣ)² - 2ˣ = 0
==> (3ˣ - 2ˣ)(3ˣ + 2ˣ) + (3ˣ - 2ˣ) = 0
==> (3ˣ - 2ˣ)(3ˣ + 2ˣ + 1) = 0
Since 3ˣ > 0 and 2ˣ > 0, so (3ˣ + 2ˣ + 1) > 0, therefore
3ˣ - 2ˣ = 0
==> 3ˣ = 2ˣ
==> (3/2)ˣ = 1
==> x = 0

9ˣ + 3ˣ = 4ˣ + 2ˣ
3ˣ(3ˣ+1) = 2ˣ(2ˣ+1)
3ˣ/2ˣ = (2ˣ+1)/(3ˣ+1)

hence we have the simultaneous equations:

3ˣ = k(2ˣ+1) -- (1)
2ˣ = k(3ˣ+1) -- (2)

where k is a positive real number since aˣ > 0 for all real number a and x

substitute (1) into (2) yields

2ˣ = k[k(2ˣ+1)+1]
2ˣ = (k^2)*2ˣ +(k^2)+k
[(k^2) -1]2ˣ +(k^2)+k = 0
(k+1)(k-1)(2ˣ) + k(k+1) = 0
(k+1)[(k-1)(2ˣ) + k] = 0 (Note: k is a positive real number, so (k+1)>0)
(k-1)(2ˣ) + k = 0
k(2ˣ) + k = 2ˣ
3ˣ=2ˣ
xln3 = xln2
x=0

hope this is correct and can help you

2013-01-08 15:02:19 補充：
NOTE: log(A+B) is not equivalent to log(A) + log(B),
therefore, the conversion from
log (9^x+3^x)=log (4^x+2^x)

to
xlog 9+xlog 3=xlog 4+xlog 2

is wrong

9^x+3^x=4^x+2^x
log (9^x+3^x)=log (4^x+2^x)
xlog 9+xlog 3=xlog 4+xlog 2
x(log 9+log 3)=x(log 4+log 2)
obviously only x=0 is the solution