Answer:
x = 5
Step-by-step explanation:
To solve the equation [tex] 9^{x+3} = 81^{x-1} [/tex], let's start by expressing [tex] 81 [/tex] as a power of [tex] 9 [/tex], since [tex] 81 = 9^2 [/tex].
Therefore, we can rewrite [tex] 81^{x-1} [/tex] as [tex] (9^2)^{x-1} [/tex]. Using the property of exponents [tex](a^m)^n = a^{m \cdot n}[/tex], we have:
[tex] (9^2)^{x-1} = 9^{2(x-1)} [/tex]
Now substitute this back into the original equation:
[tex] 9^{x+3} = 9^{2(x-1)} [/tex]
Since the bases are the same (both are [tex]9[/tex]), we can equate the exponents:
[tex] x+3 = 2(x-1) [/tex]
Now, solve for [tex] x [/tex]:
[tex] x + 3 = 2x - 2 [/tex]
Subtract [tex] x [/tex] from both sides:
[tex] 3 = x - 2 [/tex]
Add [tex] 2 [/tex] to both sides:
[tex] x = 3 + 2 [/tex]
[tex] x = 5 [/tex]
Therefore, the solution to the equation is [tex] x = 5 [/tex].
