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Fill in the missing values to make the equations true.

(a) [tex]\log _5 7+\log _5 3=\log _5 \square[/tex]

(b) [tex]\log _3 11-\log _3 \square=\log _3 \frac{11}{2}[/tex]

(c) [tex]\log _6 25=2 \log _6 \square[/tex]

Sagot :

Sure, let's fill in the missing values for these logarithmic equations one by one.

### Part (a):

We start with the expression:
[tex]\[ \log _5 7+\log _5 3=\log _5 \square \][/tex]

According to the property of logarithms, [tex]\(\log_b(m) + \log_b(n) = \log_b(m \times n)\)[/tex].

Therefore:
[tex]\[ \log _5 7 + \log _5 3 = \log _5 (7 \times 3) \][/tex]
[tex]\[ 7 \times 3 = 21 \][/tex]

So, the missing value is:
[tex]\[ \log _5 7+\log _5 3=\log _5 21 \][/tex]

### Part (b):

Next, we have the expression:
[tex]\[ \log _3 11-\log _3 \square=\log _3 \frac{11}{2} \][/tex]

According to the property of logarithms, [tex]\(\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)\)[/tex].

Therefore:
[tex]\[ \log _3 11 - \log _3 \square = \log _3 \left(\frac{11}{2}\right) \][/tex]

For these logarithmic sides to be equal, [tex]\(\square\)[/tex] must be equal to:
[tex]\[ \frac{11}{\left(\frac{11}{2}\right)} \][/tex]
[tex]\[ \frac{11}{1} \cdot \frac{2}{11} = 2 \][/tex]

So, the missing value is:
[tex]\[ \log _3 11 - \log _3 2 = \log _3 \frac{11}{2} \][/tex]

### Part (c):

Finally, we have the expression:
[tex]\[ \log _6 25 = 2 \log _6 \square \][/tex]

According to the property of logarithms, [tex]\(n \log_b(m) = \log_b(m^n)\)[/tex].

Applying this property to the given expression:
[tex]\[ \log _6 25 = 2 \log _6 \square \][/tex]

Rewriting:
[tex]\[ \log _6 25 = \log _6 (\square)^2 \][/tex]

For these to be equal:
[tex]\[ (\square)^2 = 25 \][/tex]
[tex]\[ \square = \sqrt{25} \][/tex]
[tex]\[ \square = 5 \][/tex]

So, the missing value is:
[tex]\[ \log _6 25 = 2 \log _6 5 \][/tex]

### Summary:
The filled-in equations are:
(a) [tex]\[ \log _5 7+\log _5 3=\log _5 21 \][/tex]
(b) [tex]\[ \log _3 11-\log _3 2=\log _3 \frac{11}{2} \][/tex]
(c) [tex]\[ \log _6 25=2 \log _6 5 \][/tex]