lg7(x-1)+lg7x=lg7x

To solve the equation \( \log_7(x-1) + \log_7(x) = \log_7(x) \), you can start by using the properties of logarithms.

First, observe that you can subtract \( \log_7(x) \) from both sides:

\[ \log_7(x-1) = \log_7(x) - \log_7(x) \]

This simplifies to:

\[ \log_7(x-1) = 0 \]

Next, recall that \( \log_b(a) = 0 \) implies \( a = 1 \). Therefore,

\[ x - 1 = 1 \]

Solving for \( x \):

\[ x = 2 \]

Now, you should check if this solution is valid by substituting \( x = 2 \) back into the original equation:

\[ \log_7(2-1) + \log_7(2) = \log_7(2) \]

This becomes:

\[ \log_7(1) + \log_7(2) = \log_7(2) \]

Since \( \log_7(1) = 0 \),

\[ 0 + \log_7(2) = \log_7(2) \]

Thus, both sides of the equation are equal, confirming that the solution is valid.

The final solution is:

\[ \boxed{2} \]