Question

The solution of \( 3^{2x-1} = 81^{1-x} \) is:

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{7}{6} \)
• D. \( \frac{5}{6} \)
• E. \( \frac{1}{3} \)

Hint:

Always look to simplify bases to their smallest prime factors (like 2, 3, or 5). This makes equating exponents straightforward and reduces the need for logarithms in simple exponential problems.

Answer 1

The Correct Option is D

Concept: To solve an exponential equation where the bases are different, the most effective strategy is to express both sides of the equation using the same base. Once the bases are equal (\( b^M = b^N \)), we can simply equate the exponents (\( M = N \)) and solve for the variable.

Step 1: Expressing both sides with base 3.
The left side is already in base 3. For the right side, we know that \( 81 = 3^4 \). Substituting this into the equation: \[ 3^{2x-1} = (3^4)^{1-x} \] Using the power of a power rule \( (a^m)^n = a^{m \times n} \), we simplify the right side: \[ 3^{2x-1} = 3^{4(1-x)} \] \[ 3^{2x-1} = 3^{4-4x} \]

Step 2: Equating the exponents and solving for \( x \).
Since the bases are now identical, we set the exponents equal to each other: \[ 2x - 1 = 4 - 4x \] Now, group the \( x \) terms on one side and the constants on the other: \[ 2x + 4x = 4 + 1 \] \[ 6x = 5 \] \[ x = \frac{5}{6} \]