Question

Solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer.
log left parenthesis x plus 4 right parenthesis minus log 2 equals log left parenthesis 5 x plus 1 right parenthesis
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Part 1
Solve the equation to find the solution set. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet .
  
​(Simplify your answer. Use a comma to separate answers as​ needed.)
B.
There are infinitely many solutions.
C.
There is no solution.

Answers

GPT-4o mini
To solve the equation

\[
\log(x + 4) - \log(2) = \log(5x + 1),
\]

we can apply the properties of logarithms. Specifically, we can use the property that states \(\log(a) - \log(b) = \log\left(\frac{a}{b}\right)\). This gives us

\[
\log\left(\frac{x + 4}{2}\right) = \log(5x + 1).
\]

Since both sides of the equation are logarithms, we can eliminate the logarithms by setting the arguments equal to each other, provided that the arguments are positive:

\[
\frac{x + 4}{2} = 5x + 1.
\]

Next, we will multiply both sides by 2 to eliminate the denominator:

\[
x + 4 = 10x + 2.
\]

Now, we can rearrange the equation to isolate \(x\):

\[
4 - 2 = 10x - x,
\]

which simplifies to:

\[
2 = 9x.
\]

Dividing both sides by 9 gives:

\[
x = \frac{2}{9}.
\]

Now, we need to check if this solution is in the domain of the original logarithmic expression. The arguments of the logarithms must be positive:

1. For \(\log(x + 4)\):
\[
x + 4 > 0 \implies \frac{2}{9} + 4 > 0 \text{ (which is true)}.
\]

2. For \(\log(2)\):
\(\log(2)\) is defined as 2 is a positive number.

3. For \(\log(5x + 1)\):
\[
5x + 1 > 0 \implies 5\left(\frac{2}{9}\right) + 1 = \frac{10}{9} + 1 = \frac{10}{9} + \frac{9}{9} = \frac{19}{9} > 0 \text{ (which is true)}.
\]

Since \(x = \frac{2}{9}\) is in the domain of the logarithmic functions, it is a valid solution.

Thus, the solution set is:

\[
\boxed{\left\{\frac{2}{9}\right\}}.
\]