Question

fill in the missing values to make the equations true.\\(\log_{3}7 + \log_{3}5 = \log_{3}\square\\)\\(\log_{2}7 - \log_{2}\square = \log_{2}\frac{7}{3}\\)\\(\log_{8}\frac{1}{25} = -2\log_{8}\square\\)

Explanation:

Step1: Apply log addition rule

The logarithmic addition rule is $\log_b x + \log_b y = \log_b (xy)$. For $\log_3 7 + \log_3 5$, we multiply the arguments: $7\times5 = 35$. So $\log_3 7 + \log_3 5 = \log_3 35$.

Step2: Apply log subtraction rule

The logarithmic subtraction rule is $\log_b x - \log_b y = \log_b \frac{x}{y}$. For $\log_2 7 - \log_2 \square = \log_2 \frac{7}{3}$, we have $\frac{7}{\square} = \frac{7}{3}$, so $\square = 3$.

Step3: Apply log power rule

The logarithmic power rule is $n\log_b x = \log_b x^n$ (and also $\log_b x^n = n\log_b x$). For $\log_8 \frac{1}{25} = -2\log_8 \square$, rewrite the left side: $\log_8 \frac{1}{25} = \log_8 25^{-1} = - \log_8 25$. Then $- \log_8 25 = -2\log_8 \square$ implies $\log_8 25 = 2\log_8 \square$. Using the power rule, $2\log_8 \square = \log_8 \square^2$. So $\square^2 = 25$, and since we take the positive root (as the argument of a log must be positive), $\square = 5$.

Answer:

First equation: 35; Second equation: 3; Third equation: 5