9^(2x) / 3^x = 27 × (1/81)^x

Question

ScanSolve AI Tutor · Accepted Answer

The problem asks to solve the exponential equation:

\frac{9^{2x}}{3^{0x}} = 27 \times \left(\frac{1}{81}\right)^x

Simplify the term $3^{0x}$ .
Any non-zero number raised to the power of 0 is 1. Since $3 \neq 0$ , $3^{0x} = 3^0 = 1$ .
The equation becomes:
$9^{2x} = 27 \times \left(\frac{1}{81}\right)^x$
Express all terms with a common base.
The common base here is 3.
- $9 = 3^2$
- $27 = 3^3$
- $81 = 3^4$
Substitute these into the equation:
$(3^2)^{2x} = 3^3 \times \left(\frac{1}{3^4}\right)^x$
Apply the exponent rules $(a^m)^n = a^{mn}$ and $\frac{1}{a^m} = a^{-m}$ .
$3^{2 \times 2x} = 3^3 \times (3^{-4})^x$ $3^{4x} = 3^3 \times 3^{-4x}$
Apply the exponent rule $a^m \times a^n = a^{m+n}$ on the right side.
$3^{4x} = 3^{3 + (-4x)}$ $3^{4x} = 3^{3 - 4x}$
Since the bases are equal, the exponents must be equal.
$4x = 3 - 4x$
Solve the linear equation for $x$ .
Add $4x$ to both sides:
$4x + 4x = 3$ $8x = 3$
Divide by 8:
$x = \frac{3}{8}$
The solution is $x = \frac{3}{8}$ .
$\frac{3}{8}$

9^(2x) / 3^x = 27 * (1/81)^x