Write as a Single Logarithm log base 3 of 4+ log base 3 of y+1/2* log base 3 of 49

$\log_{3} (4) + \log_{3} (y) + \frac{1}{2} \cdot \log_{3} (49)$

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{3} (4 y) + \frac{1}{2} \cdot \log_{3} (49)$

Simplify each term.

Simplify $\frac{1}{2} \log_{3} (49)$ by moving $\frac{1}{2}$ inside the logarithm.

$\log_{3} (4 y) + \log_{3} (49^{\frac{1}{2}})$

Rewrite $49$ as $7^{2}$ .

$\log_{3} (4 y) + \log_{3} ({(7^{2})}^{\frac{1}{2}})$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\log_{3} (4 y) + \log_{3} (7^{2 (\frac{1}{2})})$

Cancel the common factor of $2$ .

Cancel the common factor.

$\log_{3} (4 y) + \log_{3} (7^{2 (\frac{1}{2})})$

Rewrite the expression.

$\log_{3} (4 y) + \log_{3} (7^{1})$

Evaluate the exponent.

$\log_{3} (4 y) + \log_{3} (7)$

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{3} (4 y \cdot 7)$

Multiply $7$ by $4$ .

$\log_{3} (28 y)$