Question
Solve the equation
Log3^x= 9 log x^3
Log3^x= 9 log x^3
Answers
GPT-4o mini
To solve the equation \( \log_3 x = 9 \log x^3 \), we can simplify and rearrange the equation step by step.
First, recognize that:
\[
\log x^3 = 3 \log x
\]
So we can rewrite the equation:
\[
\log_3 x = 9 \cdot 3 \log x
\]
which simplifies to:
\[
\log_3 x = 27 \log x
\]
Next, we can use the change of base formula for logarithms, which states that \( \log_a b = \frac{\log b}{\log a} \). Applying this to \( \log_3 x \), we get:
\[
\log_3 x = \frac{\log x}{\log 3}
\]
Substituting back into our equation gives:
\[
\frac{\log x}{\log 3} = 27 \log x
\]
Now, if \( \log x \neq 0 \) (which we assume as \( x > 1 \)), we can divide both sides by \( \log x \):
\[
\frac{1}{\log 3} = 27
\]
From this, we can rearrange and solve for \( \log 3 \):
\[
\log 3 = \frac{1}{27}
\]
This equality indicates actual numbers, however, it suggests there might have been an error in handling or an inconsistency as no \( x \) satisfies this statement traditionally with the logarithm definitions.
To summarize, we re-check:
1. \( \log_3 x = \frac{\log x}{\log 3} \rightarrow \frac{1}{\log 3} = 27\) leads to an inconsistency with standard logarithm definitions.
2. Must assume: \( \log x = 0\) giving \( x = 1\) is matched leading also to check back.
Thus ensuring back we can cross-validate if \(x = 1\) satisfies original formed:
Indeed both sides yield:
LHS:
\[
\log_3 1 = 0
\]
RHS:
\[
9 \cdot 3 \cdot 0 = 0
\]
Thus, the only solution for the equation \( \log_3 x = 9 \log x^3 \) is:
\[
\boxed{1}
\]
First, recognize that:
\[
\log x^3 = 3 \log x
\]
So we can rewrite the equation:
\[
\log_3 x = 9 \cdot 3 \log x
\]
which simplifies to:
\[
\log_3 x = 27 \log x
\]
Next, we can use the change of base formula for logarithms, which states that \( \log_a b = \frac{\log b}{\log a} \). Applying this to \( \log_3 x \), we get:
\[
\log_3 x = \frac{\log x}{\log 3}
\]
Substituting back into our equation gives:
\[
\frac{\log x}{\log 3} = 27 \log x
\]
Now, if \( \log x \neq 0 \) (which we assume as \( x > 1 \)), we can divide both sides by \( \log x \):
\[
\frac{1}{\log 3} = 27
\]
From this, we can rearrange and solve for \( \log 3 \):
\[
\log 3 = \frac{1}{27}
\]
This equality indicates actual numbers, however, it suggests there might have been an error in handling or an inconsistency as no \( x \) satisfies this statement traditionally with the logarithm definitions.
To summarize, we re-check:
1. \( \log_3 x = \frac{\log x}{\log 3} \rightarrow \frac{1}{\log 3} = 27\) leads to an inconsistency with standard logarithm definitions.
2. Must assume: \( \log x = 0\) giving \( x = 1\) is matched leading also to check back.
Thus ensuring back we can cross-validate if \(x = 1\) satisfies original formed:
Indeed both sides yield:
LHS:
\[
\log_3 1 = 0
\]
RHS:
\[
9 \cdot 3 \cdot 0 = 0
\]
Thus, the only solution for the equation \( \log_3 x = 9 \log x^3 \) is:
\[
\boxed{1}
\]