Questions: Solve 1/81^(x-2) = 27^(1-x)

Solution Steps

To solve the equation \(\frac{1}{81^{(x-2)}}=27^{(1-x)}\), we can start by expressing both sides of the equation with the same base. Notice that \(81\) and \(27\) are powers of \(3\), specifically \(81 = 3^4\) and \(27 = 3^3\). Rewrite the equation using these bases and then equate the exponents to solve for \(x\).

We start with the equation

\[ \frac{1}{81^{(x-2)}} = 27^{(1-x)}. \]

Rewriting \(81\) and \(27\) in terms of base \(3\), we have

\[ 81 = 3^4 \quad \text{and} \quad 27 = 3^3. \]

Thus, the equation becomes

\[ \frac{1}{(3^4)^{(x-2)}} = (3^3)^{(1-x)}. \]

This simplifies to

\[ 3^{-4(x-2)} = 3^{3(1-x)}. \]

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

\[ -4(x-2) = 3(1-x). \]

Expanding both sides gives:

\[ -4x + 8 = 3 - 3x. \]

Rearranging the equation leads to:

\[ -4x + 3x = 3 - 8, \]

which simplifies to

\[ -x = -5. \]

Thus, we find

\[ x = 5. \]

Final Answer