Log3(5-4x)<log3(x-1)

To solve the inequality \( \log_3(5 - 4x) < \log_3(x - 1) \), we can start by using the properties of logarithms. Specifically, since the logarithmic function is increasing when the base is greater than 1, we can simplify the inequality as follows:

\[ 5 - 4x < x - 1 \]

Next, let's solve this inequality step by step:

1. Rearrange the terms:

\[ 5 + 1 < x + 4x \]

This simplifies to:

\[ 6 < 5x \]

2. Divide both sides by 5:

\[ \frac{6}{5} < x \]

Or:

\[ x > \frac{6}{5} \]

Now, we must also consider the conditions under which the logarithms are defined. The arguments of the logarithms must be positive:

1. From \( \log_3(5 - 4x) \), we need: \[ 5 - 4x > 0 \implies 4x < 5 \implies x < \frac{5}{4} \]

2. From \( \log_3(x - 1) \), we need: \[ x - 1 > 0 \implies x > 1 \]

Now we have three inequalities to satisfy:

1. \( x > \frac{6}{5} \)
2. \( x < \frac{5}{4} = 1.25 \)
3. \( x > 1 \)

Considering the inequalities \( x > \frac{6}{5} \) and \( x < \frac{5}{4} \), the solution will be between the two:

\[ \frac{6}{5} < x < \frac{5}{4} \]

In interval notation, the solution can be expressed as:

\[ \left( \frac{6}{5}, \frac{5}{4} \right) \]

So the final solution is:

\[ x \in \left( \frac{6}{5}, \frac{5}{4} \right) \]