We'll move all terms to one
side:
3*4^x + 2*9^x - 5*6^x=
0
We remark that 6^x =
(2*3)^x
But (2*3)^x =
2^x*3^x
We'll divide by 3^2x all
over:
3*(2/3)^2x - 5*(2/3)^x + 2 =
0
We'll note the exponential ratio (2/3)^x =
t
We'll square raise both sides and we'll
get:
(2/3)^2x = t^2
We'll
re-write the equation, changing the variable:
3t^2 - 5t + 2
= 0
We'll apply quadratic
formula:
t1 =
[5+sqrt(25-24)]/6
t1 =
(5+1)/6
t1 = 1
t2 =
(5-1)/6
t2 = 2/3
Now, we'll
put (2/3)^x = t1:
(2/3)^x =
1
We'll write 1 =
(2/3)^0
(2/3)^x =
(2/3)^0
Since the bases are matching, we'll apply one to
one property of exponentials:
x =
0
(2/3)^x = t2
(2/3)^x =
2/3
x = 1
The
solutions of the equation are {0 ; 1}.
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