Step 1: Understanding the Concept:
We are presented with a logarithmic equation containing the variable $x$ both inside and outside the log arguments. The strategy is to gather all terms involving $x$ on one side and manipulate them using logarithm rules to consolidate everything into a single logarithmic expression, which can then be converted into a standard algebraic or exponential equation.
Step 2: Key Formula or Approach:
1. Subtraction property of logs: $\log_b m - \log_b n = \log_b(m/n)$.
2. Power property of logs: $n \log_b m = \log_b(m^n)$.
3. Base conversion trick: Recognize that the standalone $x$ can be written as $x \log_{15} 15$.
Step 3: Detailed Explanation:
The given equation is:
\[ x + \log_{15}(5 + 3^x) = x \log_{15} 5 + \log_{15} 24 \]
Group the terms with a solitary $x$ factor together on the left side:
\[ x - x \log_{15} 5 + \log_{15}(5 + 3^x) = \log_{15} 24 \]
Factor out the $x$:
\[ x(1 - \log_{15} 5) + \log_{15}(5 + 3^x) = \log_{15} 24 \]
Since we are working with base 15 logarithms, rewrite $1$ as $\log_{15} 15$:
\[ x(\log_{15} 15 - \log_{15} 5) + \log_{15}(5 + 3^x) = \log_{15} 24 \]
Use the quotient rule for logarithms on the term inside the parentheses:
\[ x(\log_{15} (15/5)) + \log_{15}(5 + 3^x) = \log_{15} 24 \]
\[ x \log_{15} 3 + \log_{15}(5 + 3^x) = \log_{15} 24 \]
Use the power rule for logarithms to move the $x$ inside:
\[ \log_{15} (3^x) + \log_{15}(5 + 3^x) = \log_{15} 24 \]
Now, apply the product rule for logarithms ($\log a + \log b = \log ab$):
\[ \log_{15} [3^x \cdot (5 + 3^x)] = \log_{15} 24 \]
Since the logarithmic function is one-to-one, we can equate the arguments:
\[ 3^x(5 + 3^x) = 24 \]
This is a quadratic equation hidden in exponential form. Let's make a substitution: let $y = 3^x$. Note that $y$ must be strictly positive since $3^x>0$ for all real $x$.
\[ y(5 + y) = 24 \]
\[ y^2 + 5y - 24 = 0 \]
Factor the quadratic equation:
\[ (y + 8)(y - 3) = 0 \]
This yields two possible values for $y$: $y = -8$ or $y = 3$.
Since we established $y = 3^x>0$, the solution $y = -8$ is rejected.
So, we have:
\[ y = 3 \implies 3^x = 3^1 \]
Equating exponents:
\[ x = 1 \]
Step 4: Final Answer:
The value of $x$ is 1.