To solve the equation [tex]\(\frac{9^x \times 3^5 \times 27^3}{3 \times 81^4} = 27\)[/tex] for [tex]\(x\)[/tex], let's follow these steps:
1. Express all terms with base 3:
- [tex]\(9 = 3^2\)[/tex]
- [tex]\(27 = 3^3\)[/tex]
- [tex]\(81 = 3^4\)[/tex]
So, the equation becomes:
[tex]\[
\frac{(3^2)^x \times 3^5 \times (3^3)^3}{3 \times (3^4)^4} = 27.
\][/tex]
2. Simplify the exponents:
- [tex]\((3^2)^x = 3^{2x}\)[/tex]
- [tex]\( (3^3)^3 = 3^{9}\)[/tex]
- [tex]\( (3^4)^4 = 3^{16}\)[/tex]
Substituting these into the equation, we get:
[tex]\[
\frac{3^{2x} \times 3^5 \times 3^9}{3 \times 3^{16}} = 27.
\][/tex]
3. Combine the exponents in the numerator:
Adding exponents in the numerator:
[tex]\[
3^{2x} \times 3^5 \times 3^9 = 3^{2x + 5 + 9} = 3^{2x + 14}.
\][/tex]
So now, the equation is:
[tex]\[
\frac{3^{2x + 14}}{3 \times 3^{16}} = 27.
\][/tex]
4. Subtract the exponent in the denominator:
[tex]\[
\frac{3^{2x + 14}}{3^1 \times 3^{16}} = 3^{2x + 14 - 1 - 16} = 3^{2x - 3}.
\][/tex]
Therefore, we have:
[tex]\[
3^{2x - 3} = 27.
\][/tex]
5. Express 27 as a power of 3:
[tex]\[
27 = 3^3.
\][/tex]
So the equation now is:
[tex]\[
3^{2x - 3} = 3^3.
\][/tex]
6. Since the bases are the same, set the exponents equal to each other:
[tex]\[
2x - 3 = 3.
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[
2x = 3 + 3,
\][/tex]
[tex]\[
2x = 6,
\][/tex]
[tex]\[
x = \frac{6}{2},
\][/tex]
[tex]\[
x = 3.
\][/tex]
Thus, the value of [tex]\(x\)[/tex] is [tex]\(3\)[/tex].
