Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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.