Sure, let's go through each equation step-by-step.
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### Part (a)
Given equation:
[tex]\[ \log_3 2 + \log_3 7 = \log_3 \text{!} \][/tex]
Using the properties of logarithms, specifically the addition property which states [tex]\( \log_b(a) + \log_b(c) = \log_b(ac) \)[/tex]:
[tex]\[ \log_3 2 + \log_3 7 = \log_3 (2 \times 7) \][/tex]
So:
[tex]\[ \log_3 2 + \log_3 7 = \log_3 14 \][/tex]
Thus, the missing value is 14:
[tex]\[ \boxed{14} \][/tex]
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### Part (b)
Given equation:
[tex]\[ \log_2 5 - \log_2 \square = \log_2 \frac{5}{3} \][/tex]
Using the properties of logarithms, specifically the subtraction property which states [tex]\( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \)[/tex]:
[tex]\[ \log_2 5 - \log_2 \text{!} = \log_2 \left(\frac{5}{x}\right) \][/tex]
For the equation to hold true as given:
[tex]\[ \log_2 5 - \log_2 x = \log_2 \frac{5}{3} \][/tex]
It must be that:
[tex]\[ x = 3 \][/tex]
Thus, the missing value is 3:
[tex]\[ \boxed{3} \][/tex]
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### Part (c)
Given equation:
[tex]\[ \log_6 \frac{1}{32} = -5 \log_6 \square \][/tex]
Using the properties of logarithms, specifically the power rule which states [tex]\( \log_b(a^c) = c \log_b(a) \)[/tex]:
[tex]\[ \log_6 \frac{1}{32} = -5 \log_6 x \][/tex]
We can rewrite the right-hand side:
[tex]\[ \log_6 \frac{1}{32} = \log_6 (x^{-5}) \][/tex]
For the equation to hold true, we need:
[tex]\[ x^{-5} = \frac{1}{32} \][/tex]
Taking the reciprocal and solving for [tex]\( x \)[/tex]:
[tex]\[ x^5 = 32 \][/tex]
So:
[tex]\[ x = 2 \][/tex]
Thus, the missing value is 2:
[tex]\[ \boxed{2} \][/tex]
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Therefore, the missing values are:
(a) 14
(b) 3
(c) 2
