Answer:
x = 1
Step-by-step explanation:
Given equation:
[tex]9^{x+2}=\left(\dfrac{1}{27}\right)^{x-3}[/tex]
To solve the equation for x, begin by expressing the bases as powers of the same base.
Both 9 and 27 can be written as powers of 3, so:
[tex]\left(3^2\right)^{x+2}=\left(\dfrac{1}{3^3}\right)^{x-3}[/tex]
Apply the negative exponent rule, which states that any non-zero base raised to a negative exponent is equal to the reciprocal of the base raised to the positive exponent:
[tex]\left(3^2\right)^{x+2}=\left(3^{-3}\right)^{x-3}[/tex]
Now, apply the power of a power rule, which states that when raising a power to another power, we multiply the exponents:
[tex]3^{2(x+2)}=3^{-3(x-3)}\\\\\\3^{2x+4}=3^{-3x+9}[/tex]
Since the bases are the same, the exponents must be equal, so:
[tex]2x+4=-3x+9[/tex]
Solve for x:
[tex]2x+4+3x=-3x+9+3x\\\\5x+4=9\\\\5x+4-4=9-4\\\\5x=5\\\\\dfrac{5x}{5}=\dfrac{5}{5}\\\\x=1[/tex]
Therefore, the value of x is:
[tex]\LARGE\boxed{\boxed{x=1}}[/tex]
