Use
the rules for exponents to restate in the manner outlined in a previous reply:
. . . . .$\displaystyle 9^{x+1}\, =\, 9^x\, \times\, 9^1\, =\, (9^x)(9)\, =\, 9\, \times\, 9^x$
. . . . .$\displaystyle 3^{x+2}\, =\, 3^x\, \times\, 3^2\, =\, 3^x\, \times\, 9\, \, 9\, \times\, 3^x$
This makes the equation the same as:
. . . . .$\displaystyle 9(9^x)\, +\, 9(3^x)\, -\, 4\, =\, 0$
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:
. . . . .$\displaystyle 9(3^x)^2\, +\, 9(3^x)\, -\, 4\, =\, 0$
Using the factoring provided earlier, you should arrive at:
. . . . .$\displaystyle 3^x\, =\, -\frac{4}{3}\, \mbox{ and }\, 3^x\, =\, \frac{1}{3}$
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::
. . . . .$\displaystyle x\, +\, 1\, =\, \log_3(4/3)\, +\, \log_3(3)\, =\, \log_3(4)$
. . . . .$\displaystyle x\, +\, 2\, =\, \log_3(4/3)\, +\, \log_3(9)\, =\, \log_3(8)$
. . . . .$\displaystyle 9^{x+1}\, =\, \left(3^{\log_3(4)}\right)^2\, =\, (4)^2\, =\, 16$
. . . . .$\displaystyle 3^{x+2}\, =\, 3^{\log_3(8)}\, =\, 8$
. . . . .$\displaystyle 9^{x+1}\, +\, 3^{x+2}\, -\, 4\, =\, 16\, +\, 8\, -\, 4\, =\, 20$
But the left-hand side is
supposed to equal zero.
You might want to point out this typo to your instructor.