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Exponentials and Logarithms

Basic properties of logarithms

Fill in the missing values to make the equations true.

(a) [tex]\(\log_7 3 - \log_7 8 = \log_7 \square\)[/tex]

(b) [tex]\(\log_5 2 + \log_5 \square = \log_5 14\)[/tex]

(c) [tex]\(3 \log_3 2 = \log_3 \square\)[/tex]


Sagot :

Certainly! Let's solve each of these logarithmic equations step-by-step:

### (a) [tex]\(\log_7 3 - \log_7 8 = \log_7 \square\)[/tex]

We use the property of logarithms that states:
[tex]\[ \log_b a - \log_b c = \log_b \left(\frac{a}{c}\right) \][/tex]
Applying this property, we have:
[tex]\[ \log_7 3 - \log_7 8 = \log_7 \left(\frac{3}{8}\right) \][/tex]

Thus, the missing value in the equation is:
[tex]\[ \square = \frac{3}{8} \][/tex]

So, (a) [tex]\(\log_7 3 - \log_7 8 = \log_7 \left(\frac{3}{8}\right)\)[/tex].


### (b) [tex]\(\log_5 2 + \log_5 \square = \log_5 14\)[/tex]

We use the property of logarithms that states:
[tex]\[ \log_b a + \log_b c = \log_b (a \cdot c) \][/tex]
Applying this property, we have:
[tex]\[ \log_5 2 + \log_5 \square = \log_5 (2 \cdot \square) \][/tex]
Given that it equals [tex]\(\log_5 14\)[/tex], we can equate:
[tex]\[ 2 \cdot \square = 14 \][/tex]
Solving for the missing value, we get:
[tex]\[ \square = \frac{14}{2} = 7 \][/tex]

So, (b) [tex]\(\log_5 2 + \log_5 7 = \log_5 14\)[/tex].


### (c) [tex]\(3 \log_3 2 = \log_3 \square\)[/tex]

We use the property of logarithms that states:
[tex]\[ n \log_b a = \log_b (a^n) \][/tex]
Applying this property, we have:
[tex]\[ 3 \log_3 2 = \log_3 (2^3) \][/tex]

Thus, the missing value in the equation is:
[tex]\[ \square = 2^3 = 8 \][/tex]

So, (c) [tex]\(3 \log_3 2 = \log_3 8\)[/tex].


### Summary of Solutions:
(a) [tex]\(\log_7 3 - \log_7 8 = \log_7 \left(\frac{3}{8}\right)\)[/tex]
(b) [tex]\(\log_5 2 + \log_5 7 = \log_5 14\)[/tex]
(c) [tex]\(3 \log_3 2 = \log_3 8\)[/tex]

We have filled in the missing values to make the equations true.
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