Verified Answer :

\((2^-^5) ^x = 2 ^ - ^1^5 \\ \\ ( \frac{1}{32} ) ^x = \frac{1}{32768} \\ \\ \hbox{Solve the exponent.} \\ \\ ( \frac{1}{32})^x = \frac{1}{32768} \\ \\\hbox{Take log of both sides} \\ \\ \hbox{log}} ( \frac{1}{32}) ^ x = log ( \frac{1}{32768}) \\ \\ x \times log (\frac{1}{32} ) = log ( \frac{1}{32768} ) \\ \\ \bold{ \frac{log( \frac{1}{32768})}{log \frac{1}{32} } } \\ \\ x = 3\)

Therefore the missing exponent is 3. Which makes the equation true.

\({ \left( { 2 }^{ -5 } \right) }^{ x }={ 2 }^{ -15 }\\ \\ { 2 }^{ -5x }={ 2 }^{ -15 }\\ \\ \therefore \quad -5x=-15\\ \\ \therefore \quad x=3\)

Remember that:

\({ \left( { a }^{ m } \right) }^{ n }={ a }^{ m\cdot n }\)

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