Question

Solve the logarithmic equation log4 (6−n)=log4 (3n)

Answer 1

To solve the equation \( \log_4 (6 - n) = \log_4 (3n) \), we can use the property that if \( \log_a (x) = \log_a (y) \), then \( x = y \) (assuming \( a > 0 \) and \( a \neq 1 \)).

Thus, we can set the arguments of the logarithms equal to each other:

\[
6 - n = 3n
\]

Now, we can solve this equation for \( n \):

1. Add \( n \) to both sides:

\[
6 = 3n + n
\]

2. Combine like terms:

\[
6 = 4n
\]

3. Divide both sides by 4:

\[
n = \frac{6}{4} = \frac{3}{2}
\]

Now, we need to check if this value of \( n \) is valid by substituting it back into the original logarithmic arguments to ensure they are positive:

1. For \( 6 - n \):

\[
6 - \frac{3}{2} = \frac{12}{2} - \frac{3}{2} = \frac{9}{2} \quad \text{(positive)}
\]

2. For \( 3n \):

\[
3n = 3 \times \frac{3}{2} = \frac{9}{2} \quad \text{(positive)}
\]

Since both arguments are positive, \( n = \frac{3}{2} \) is a valid solution.

Thus, the solution to the logarithmic equation is:

\[
\boxed{\frac{3}{2}}
\]