Questions: 4^-n/16^-n=1

4^-n/16^-n=1
Transcript text: $\frac{4^{-n}}{16^{-n}}=1$
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Solution

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Solution Steps

To solve the equation \(\frac{4^{-n}}{16^{-n}}=1\), we can simplify the expression by recognizing that \(16^{-n}\) can be rewritten as \((4^2)^{-n} = 4^{-2n}\). This allows us to rewrite the equation as \(\frac{4^{-n}}{4^{-2n}} = 1\). By simplifying the fraction, we can solve for \(n\).

Step 1: Simplify the Expression

The given equation is \(\frac{4^{-n}}{16^{-n}} = 1\). We can rewrite \(16^{-n}\) as \((4^2)^{-n} = 4^{-2n}\). This simplifies the equation to:

\[ \frac{4^{-n}}{4^{-2n}} = 1 \]

Step 2: Simplify the Fraction

The expression \(\frac{4^{-n}}{4^{-2n}}\) can be simplified using the properties of exponents:

\[ 4^{-n - (-2n)} = 4^{n} \]

Thus, the equation becomes:

\[ 4^{n} = 1 \]

Step 3: Solve the Equation

The equation \(4^{n} = 1\) implies that \(n\) must be such that \(4^{n} = 4^{0}\), since \(4^{0} = 1\). Therefore, \(n = 0\).

Additionally, considering the complex solutions, \(4^{n} = 1\) can also be expressed in terms of complex logarithms, leading to another solution:

\[ n = \frac{i\pi}{\log(2)} \]

Final Answer

The solutions to the equation \(\frac{4^{-n}}{16^{-n}} = 1\) are:

\[ \boxed{n = 0} \quad \text{and} \quad \boxed{n = \frac{i\pi}{\log(2)}} \]

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