6) \( 7^{\lg x}-5^{\lg x+1}=3 \cdot 5^{\lg x-1}-13 \cdot 7^{\lg x-1} \)

First, let's simplify the expression using the property of logarithms and exponents. We can denote \( y = \lg x \), which changes our equation to: \[ 7^y - 5^{y+1} = 3 \cdot 5^{y-1} - 13 \cdot 7^{y-1} \] This transforms the equation further into: \[ 7^y - 5 \cdot 5^y = \frac{3}{5} \cdot 5^y - \frac{13}{7} \cdot 7^y \] Now, let's combine similar terms. \[ 7^y + \frac{13}{7} \cdot 7^y = 5 \cdot 5^y + \frac{3}{5} \cdot 5^y \] Combining the coefficients yields: \[ \left(1 + \frac{13}{7}\right) 7^y = \left(5 + \frac{3}{5}\right) 5^y \] or \[ \frac{20}{7} 7^y = \frac{28}{5} 5^y \] Next, we can eliminate the fractions by multiplying both sides by \( 35 \): \[ 100 \cdot 7^y = 196 \cdot 5^y \] Now, divide both sides by \( 7^y \) to find the ratio: \[ 100 = 196 \left(\frac{5}{7}\right)^y \] Calculating \( \left(\frac{5}{7}\right)^y \) leads to: \[ \left(\frac{5}{7}\right)^y = \frac{100}{196} = \frac{25}{49} \] Taking logarithm base 10 on both sides will allow us to solve for \( y \): \[ y \cdot \lg \left(\frac{5}{7}\right) = \lg \left(\frac{25}{49}\right) \] Use the change of base property of logarithms: \[ y = \frac{\lg \left(\frac{25}{49}\right)}{\lg \left(\frac{5}{7}\right)} \] Solve for \( y \) which corresponds to \( \lg x \): \[ x = 10^{y} \] Thus, the final solution involves evaluating \( x \) from the equation derived, and substituting back through those logs.