Prove that `(log_(2)8)/(log_(3)9) =3/2`

Answer

To prove that \(\frac{\log_{2} 8}{\log_{3} 9} = \frac{3}{2}\), we can follow these steps:

### Step 1: Rewrite the logarithms
We start with the left-hand side (LHS):
\[ \frac{\log_{2} 8}{\log_{3} 9} \]

### Step 2: Express \(8\) and \(9\) in terms of powers
We know that:
\[ 8 = 2^3 \quad \text{and} \quad 9 = 3^2 \]
Thus, we can rewrite the logarithms:
\[ \log_{2} 8 = \log_{2} (2^3) \quad \text{and} \quad \log_{3} 9 = \log_{3} (3^2) \]

### Step 3: Apply the power rule of logarithms
Using the property of logarithms that states \(\log_{a} (b^n) = n \cdot \log_{a} b\), we can simplify:
\[ \log_{2} (2^3) = 3 \cdot \log_{2} 2 \quad \text{and} \quad \log_{3} (3^2) = 2 \cdot \log_{3} 3 \]

### Step 4: Substitute the simplified logarithms back into the LHS
Now substituting these back into our expression:
\[ \frac{\log_{2} 8}{\log_{3} 9} = \frac{3 \cdot \log_{2} 2}{2 \cdot \log_{3} 3} \]

### Step 5: Simplify using the fact that \(\log_{a} a = 1\)
Since \(\log_{2} 2 = 1\) and \(\log_{3} 3 = 1\), we can simplify further:
\[ \frac{3 \cdot 1}{2 \cdot 1} = \frac{3}{2} \]

### Conclusion
Thus, we have shown that:
\[ \frac{\log_{2} 8}{\log_{3} 9} = \frac{3}{2} \]
This completes the proof.

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