Question

Solve the logarithmic equation. Remember to check your solution log9 (x-5) + log9 (x+3) =1

Answer 1

To solve the logarithmic equation \( \log_9 (x-5) + \log_9 (x+3) = 1 \), we can use the properties of logarithms. Specifically, we can combine the two logarithms using the product property:

\[
\log_a b + \log_a c = \log_a (bc)
\]

Applying this property, we have:

\[
\log_9 ((x-5)(x+3)) = 1
\]

Next, we can rewrite the equation in exponential form. The logarithmic equation \( \log_9 y = 1 \) is equivalent to \( y = 9^1 \):

\[
(x-5)(x+3) = 9
\]

Now we can expand the left side:

\[
x^2 + 3x - 5x - 15 = 9
\]

This simplifies to:

\[
x^2 - 2x - 15 = 9
\]

Now we can move 9 to the left side:

\[
x^2 - 2x - 15 - 9 = 0
\]

This simplifies to:

\[
x^2 - 2x - 24 = 0
\]

Next, we can factor the quadratic:

\[
(x - 6)(x + 4) = 0
\]

Setting each factor equal to zero gives us:

\[
x - 6 = 0 \quad \text{or} \quad x + 4 = 0
\]

Thus, the solutions are:

\[
x = 6 \quad \text{or} \quad x = -4
\]

Now, we need to check our solutions to ensure they are valid in the original logarithmic equation.

### Check \( x = 6 \):

\[
\log_9(6 - 5) + \log_9(6 + 3) = \log_9(1) + \log_9(9) = 0 + 1 = 1
\]

This is valid.

### Check \( x = -4 \):

\[
\log_9(-4 - 5) + \log_9(-4 + 3) = \log_9(-9) + \log_9(-1)
\]

Since logarithms of negative numbers are undefined in the real number system, this value is not valid.

Thus, the only valid solution is:

\[
\boxed{6}
\]