Use
the rules for exponents to restate in the manner outlined in a previous reply:
. . . . . . . . . .
This makes the equation the same as:
. . . . .
Then note that 9 = 3^2, so 9^x = (3^2)^x = (3^x)^2, which means the equation is really
a "quadratic" in 3^x:
. . . . .
Using the factoring provided earlier, you should arrive at:
. . . . .
From what you've learned about
exponential functions, you know that you cannot, by powers, turn a positive three into a negative value, so the first of the two equations above has "no solution".
Apply logs or the definition to
solve the other exponential equation.
Note: The solution provided cannot be correct. If x = log_3(4/3), then
log definitions and
log rules give::
. . . . . . . . . . . . . . . . . . . . . . . . .
But the left-hand side is
supposed to equal zero.
You might want to point out this typo to your instructor.