2. Solve for the unknown variable. 2.1. \( \log 7^{2 x-1}=\log 3^{x+3} \)

We are given the equation   log(7^(2x - 1)) = log(3^(x + 3)). Since the logarithm function is one‐to‐one (provided the arguments are positive, which they are for these exponential expressions), we can set the arguments equal to each other:   7^(2x - 1) = 3^(x + 3). To solve for x, take the natural logarithm (ln) of both sides:   ln(7^(2x - 1)) = ln(3^(x + 3)). Use the logarithm power rule (ln(a^b) = b·ln(a)) on both sides:   (2x - 1) ln 7 = (x + 3) ln 3. Now, expand and rearrange the equation in order to collect the x terms on one side:   2x ln 7 - ln 7 = x ln 3 + 3 ln 3. Bring all terms involving x to one side and constant terms to the other side:   2x ln 7 - x ln 3 = ln 7 + 3 ln 3. Factor out x on the left side:   x(2 ln 7 - ln 3) = ln 7 + 3 ln 3. Finally, solve for x by dividing both sides by (2 ln 7 - ln 3):   x = (ln 7 + 3 ln 3) / (2 ln 7 - ln 3). This is the solution for x.