Write the expression as a single logarithm. \[ 7 \log _{3}(2 y+1)+\frac{1}{2} \log _{3}(y+6) \] $\log _{i]}(\square)$
Solution
Step 1 ：We are given the expression \(7 \log _{3}(2 y+1)+\frac{1}{2} \log _{3}(y+6)\).
Step 2 ：We can use the properties of logarithms to simplify this expression. The property \(k \log_b(m) = \log_b(m^k)\) allows us to move a coefficient into the argument as an exponent.
Step 3 ：Applying this property to each term, we get \(\log _{3}(2 y+1)^7+\log _{3}(y+6)^{\frac{1}{2}}\).
Step 4 ：We can then use the property \(\log_b(mn) = \log_b(m) + \log_b(n)\) to combine the two logarithms into a single logarithm of the product of the arguments.
Step 5 ：Doing so, we get \(\log _{3}\left((2 y+1)^{7}(y+6)^{\frac{1}{2}}\right)\).
Step 6 ：So, the expression \(7 \log _{3}(2 y+1)+\frac{1}{2} \log _{3}(y+6)\) simplifies to \(\boxed{\log _{3}\left((2 y+1)^{7}(y+6)^{\frac{1}{2}}\right)}\).
