Answer:
[tex]a = -4[/tex]
Step-by-step explanation:
The equation to solve for "a" is:
[tex](\frac{1}{9})^{a+1}=81^{a+1}*27^{2-a}[/tex]
The first step is to convert all of them to same bases.
9, 81, and 27, all can be expressed in base 3, lets do this:
[tex](\frac{1}{9})^{a+1}=81^{a+1}*27^{2-a}\\(3^{-2})^{a+1}=(3^4)^{a+1}*(3^3)^{2-a}[/tex]
We can use the property: [tex](a^b)^c=a^{bc}[/tex] to simplify further:
[tex](3^{-2})^{a+1}=(3^4)^{a+1}*(3^3)^{2-a}\\3^{-2a-2}=3^{4a+4}*3^{6-3a}[/tex]
The right side has 2 same bases multiplied, we can simplify this using the property: [tex]a^x * a^y = a^{x+y}[/tex]
Thus, we have:
[tex]3^{-2a-2}=3^{4a+4}*3^{6-3a}\\3^{-2a-2}=3^{4a+4+6-3a}}\\3^{-2a-2}=3^{a+10}[/tex]
Now, both sides have same base, so exponents would be equal. Now lets equate and solve for "a":
[tex]3^{-2a-2}=3^{a+10}\\-2a-2=a+10\\3a=-12\\a=-4[/tex]
So,
a = -4
