Question

Which of the following describes the solution(s) and of the equation
log
(
4
+
2
x
)
=
log
(
x
2
+
5
x
)
?(1 point)
Responses

x
=
1
is a true solution and
x
=
−
4
is an extraneous solution.
x equals 1 is a true solution and x equals negative 4 is an extraneous solution.

x
=
−
4
is a true solution and
x
=
1
is an extraneous solution.
x equals negative 4 is a true solution and x equals 1 is an extraneous solution.

x
=
1
and
x
=
−
4
are true solutions.
x equals 1 and x equals negative 4 are true solutions.

x
=
1
and
x
=
−
4
are extraneous solutions.

Answer 1

To determine the true solution(s) for the equation

\[
\log(4 + 2x) = \log(x^2 + 5x)
\]

we can start by using the property of logarithms that states if \(\log(a) = \log(b)\), then \(a = b\) given that \(a > 0\) and \(b > 0\).

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

\[
4 + 2x = x^2 + 5x
\]

Rearranging this equation gives:

\[
0 = x^2 + 5x - 2x - 4
\]

which simplifies to:

\[
0 = x^2 + 3x - 4
\]

Next, we can factor this quadratic:

\[
0 = (x + 4)(x - 1)
\]

Setting each factor to zero gives us the potential solutions:

\[
x + 4 = 0 \quad \text{or} \quad x - 1 = 0
\]

Thus, the solutions are:

\[
x = -4 \quad \text{and} \quad x = 1
\]

Now, we need to check if these solutions are valid by substituting them back into the original logarithmic expressions:

1. For \(x = 1\):
- Left side: \(\log(4 + 2(1)) = \log(6)\)
- Right side: \(\log(1^2 + 5(1)) = \log(6)\)
- Since both sides are equal, \(x = 1\) is a valid solution.

2. For \(x = -4\):
- Left side: \(\log(4 + 2(-4)) = \log(4 - 8) = \log(-4)\) (undefined)
- Right side: \(-4^2 + 5(-4) = 16 - 20 = -4\) (also undefined)
- Since the logarithm of a negative number is undefined, \(x = -4\) is not a valid solution.

Therefore, \(x = 1\) is the only true solution, and \(x = -4\) is an extraneous solution.

So the correct response is:

**x = 1 is a true solution and x = −4 is an extraneous solution.**