We are asked to solve two separate logarithm problems. Part (a) asks us to evaluate $log_3(\frac{1}{243})$. Part (b) asks us to solve the equation $log_3 81 = (2x-1)$ for $x$.

1. Problem Description

We are asked to solve two separate logarithm problems.

Part (a) asks us to evaluate $log_3(\frac{1}{243})$.

Part (b) asks us to solve the equation $log_3 81 = (2x-1)$ for $x$.

2. Solution Steps

(a) To evaluate $log_3(\frac{1}{243})$, we need to express $\frac{1}{243}$ as a power of

3. Since $243 = 3^5$, we have $\frac{1}{243} = \frac{1}{3^5} = 3^{-5}$.

Therefore, $log_3(\frac{1}{243}) = log_3(3^{-5}) = -5$.

(b) To solve the equation $log_3 81 = (2x-1)$, we need to find the value of $log_3 81$.

Since $81 = 3^4$, we have $log_3 81 = log_3(3^4) = 4$.

Thus, the equation becomes $4 = 2x - 1$.

Adding 1 to both sides gives $5 = 2x$.

Dividing both sides by 2 gives $x = \frac{5}{2} = 2.5$.

3. Final Answer

(a) -5

(b) 2.5