9514 1404 393
Answer:
y = 0
Step-by-step explanation:
Factor out 3^y:
3^(2y-1) +2(3^(y-1)) = 1 . . . . . . given
(3^y)^2(3^-1) +2(3^y)(3^-1) = 1 . . . . factor out 3^y
(3^y)^2 +2(3^y) -3 = 0 . . . . . . . . . . subtract 1, multiply by 3
Let x = 3^y. Substituting, we have ...
x^2 +2x -3 = 0
(x -1)(x +3) = 0
x = 1 or x = -3 . . . . . values of x that satisfy this equation
Then the corresponding values of y are ...
x = 3^y . . . . . . . . . . . . relation of x and y
log(x) = y·log(3) . . . . . take logs to solve for y
y = log(x)/log(3) . . . . . divide by the coefficient of y
For x = 1, this is y = log(1)/log(3) = 0
For x = -3, this is y = log(-3)/log(3) . . . . no real solutions
The only real solution is y = 0.
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The attachment shows the solution of f(x) = 0, where ...
[tex]f(x)=3^{2x-1}+2\times3^{x-1}-1[/tex]
That solution is x=0. This means the (real) solution to the given equation is y=0.
