Question

Solve the logarithmic equation.
log 4 x plus log 3 x equals log 300
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Part 1
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​(Simplify your answer. Type an exact​ answer, using e as​ needed.)

Answer 1

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;
\[
x = e^{ \left( \log_{10} 3 + 2 \right) \cdot \frac{\ln 4 \ln 3}{\ln 12}}.
\]