For what value of [tex]\(a\)[/tex] does [tex]\(\left(\frac{1}{9}\right)^2 = 81^{2+1} \cdot 27^{2-a} \)[/tex]?

A. [tex]\(-4\)[/tex]
B. [tex]\(-2\)[/tex]
C. 2
D. 6



Answer :

To solve the equation [tex]\(\left(\frac{1}{9}\right)^2 = 81^{2+1} \cdot 27^{2-a}\)[/tex], we begin by rewriting the bases in terms of powers of 3.

1. Rewriting both sides of the equation:

[tex]\[ \left(\frac{1}{9}\right)^2 = \left(\frac{1}{3^2}\right)^2 = 3^{-4} \][/tex]

[tex]\[ 81^{2+1} \cdot 27^{2-a} = 81^3 \cdot 27^{2-a} \][/tex]

2. Expressing 81 and 27 as powers of 3:

- [tex]\(81 = 3^4\)[/tex]
- [tex]\(27 = 3^3\)[/tex]

3. Substitute these powers back into the equation:

[tex]\[ 3^{-4} = (3^4)^3 \cdot (3^3)^{2-a} \][/tex]

4. Simplify the exponents in the right-hand side expression:

[tex]\[ (3^4)^3 = 3^{4 \cdot 3} = 3^{12} \][/tex]

[tex]\[ (3^3)^{2-a} = 3^{3 \cdot (2-a)} = 3^{6 - 3a} \][/tex]

So the equation becomes:

[tex]\[ 3^{-4} = 3^{12} \cdot 3^{6 - 3a} \][/tex]

5. Combine the exponents on the right-hand side:

[tex]\[ 3^{-4} = 3^{12 + 6 - 3a} = 3^{18 - 3a} \][/tex]

6. Since the bases are the same, we can equate the exponents:

[tex]\[ -4 = 18 - 3a \][/tex]

7. Solve for [tex]\(a\)[/tex]:

[tex]\[ -4 = 18 - 3a \][/tex]

Subtract 18 from both sides:

[tex]\[ -4 - 18 = -3a \][/tex]

[tex]\[ -22 = -3a \][/tex]

Divide both sides by -3:

[tex]\[ a = \frac{22}{3} \approx 7.333 \][/tex]

Thus, the value of [tex]\(a\)[/tex] is approximately [tex]\(7.333\)[/tex].

So the value of [tex]\(a\)[/tex] is not listed among the given choices ([tex]\(-4, -2, 2, 6\)[/tex]). Therefore, there seems to be a misunderstanding, but the correct mathematical value is [tex]\( \boxed{ \approx 7.333 } \)[/tex].