To solve the logarithmic equation
\[ \log_4 x + \log_3 x = \log_{10} 300, \]
we can start by converting the logarithms with bases 4 and 3 to a common base. We will use the natural logarithm (base \( e \)) for this.
Using the change of base formula, we have:
\[ \log_4 x = \frac{\ln x}{\ln 4} \quad \text{and} \quad \log_3 x = \frac{\ln x}{\ln 3}. \]
Substituting these expressions into the original equation gives:
\[ \frac{\ln x}{\ln 4} + \frac{\ln x}{\ln 3} = \log_{10} 300. \]
Now, factor out \( \ln x \) from the left side:
\[ \ln x \left( \frac{1}{\ln 4} + \frac{1}{\ln 3} \right) = \log_{10} 300. \]
Next, we simplify the left side:
\[ \ln x \left( \frac{1}{\ln 4} + \frac{1}{\ln 3} \right) = \ln x \left( \frac{\ln 3 + \ln 4}{\ln 4 \cdot \ln 3} \right) = \ln x \cdot \frac{\ln(3 \cdot 4)}{\ln 4 \cdot \ln 3} = \ln x \cdot \frac{\ln 12}{\ln 4 \cdot \ln 3}. \]
Thus, we have:
\[ \ln x \cdot \frac{\ln 12}{\ln 4 \cdot \ln 3} = \log_{10} 300. \]
Next, we need to calculate \( \log_{10} 300 \):
\[ \log_{10} 300 = \log_{10} (3 \cdot 100) = \log_{10} 3 + \log_{10} 100 = \log_{10} 3 + 2. \]
Now, substituting back, we multiply both sides by \( \frac{\ln 4 \cdot \ln 3}{\ln 12} \):
\[ \ln x = \left( \log_{10} 3 + 2 \right) \cdot \frac{\ln 4 \cdot \ln 3}{\ln 12}. \]
Therefore, to find \( x \), we exponentiate:
\[ x = e^{\left( \log_{10} 3 + 2 \right) \cdot \frac{\ln 4 \cdot \ln 3}{\ln 12}}. \]
This expression can be the exact answer in terms of \( e \). However, if you need it in another simplified format, we can approximate further using the properties of logarithms and numerical values.
If you want to compute it numerically:
- Use \( \ln 4 \approx 1.386 \)
- Use \( \ln 3 \approx 1.099 \)
- Use \( \ln 12 \approx 2.485 \)
- Use \( \log_{10} 3 \approx 0.477 \)
Thus you can calculate the numerical value of \( x \).
Hence, the final answer for \( x \) can be expressed as:
\[ x = e^{ \left( 0.477 + 2 \right) \cdot \frac{1.386 \cdot 1.099}{2.485}}. \]
For exact representation, you can type the log expressions back if required; \[ x = e^{ \left( \log_{10} 3 + 2 \right) \cdot \frac{\ln 4 \ln 3}{\ln 12}}. \]
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