To solve the equation \( \frac{e^?}{e^{65}} = \frac{e}{38} \), we can start by simplifying the left side. We can use the property of exponents that states \( \frac{e^a}{e^b} = e^{a-b} \).
Thus, we have:
\[ \frac{e^?}{e^{65}} = e^{? - 65} \]
Now, we can set the left side equal to the right side:
\[ e^{? - 65} = \frac{e}{38} \]
The right side can be rewritten using exponential notation where \( \frac{e}{38} = e^{1} \times \frac{1}{38} \). However, it is clearer to leave it as is for our purposes.
To set exponents equal (since they are both bases of \( e \)), we first rewrite the right side as:
\[ e^{1} \div e^{\log(38)} = e^{1 - \log(38)} \]
Thus we can set the exponents equal to each other:
\[ ? - 65 = 1 - \log(38) \]
Now, we can solve for \( ? \):
\[ ? = 1 - \log(38) + 65 \]
Simplifying this gives:
\[ ? = 66 - \log(38) \]
Thus, the missing exponent is \( 66 - \log(38) \). If you're looking for a numerical answer, you'll need to compute \( \log(38) \) (which is approximately \( 1.581 \)).
Therefore, \( ? \approx 66 - 1.581 \approx 64.419 \) (if calculated).
For the final answer in terms of an expression, it would be \( 66 - \log(38) \).
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