Question

Write the following expression as a single logarithm with coefficient 1.

Answer 1

Answer:

Option b.

$log_3(6c)+log_3(1/12)=log_3(c/2)$

Explanation:

To write the given expression as a single logarithm with coefficient 1 you use the properties of logaritms.

You can follow this procedure:

1) Given expression:

$log_3(6c)+log_3(1/12)$

2) Property: logarithm of a quotien

$log_b(x/y)=log_bx-log_by\\ \\ log_b1=0$

∴ $log_3(1/12)=log_31-log_312=0-log_312=-log_312$

⇒ $log_3(6c)+log_3(1/12)=log_3(6c)-log_312$

3) Property: difference of logarithms equals logarithm of the quotient

$log_3(6c)-log_312= log_3\frac{6c}{12}=log_3 \frac{c}{2}$

Which is the option B.

Answer 2

The answer is B) log3 c/2