To solve the equation \( 27^{3/4} - 1 = 26 \), we will first isolate the term involving the exponent.
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Add 1 to both sides: \[ 27^{3/4} = 27 \]
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Next, we know that \( 27 = 3^3 \), so we can rewrite \( 27^{3/4} \) as: \[ (3^3)^{3/4} = 3^{3 \cdot (3/4)} = 3^{9/4} \]
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Then we set \( 3^{9/4} = 27 \) back to \( 27 = 3^3 \): \[ 3^{9/4} = 3^3 \]
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Since the bases are the same, we can equate the exponents: \[ \frac{9}{4} = 3 \]
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Solving for \( 9 = 12 \) is incorrect, thus showing there was a misunderstanding.
The correct way to test the answer choices is by substituting each option back and checking if \( 27^{3/4} - 1 = 26 \).
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Check \( x = 1 \): \[ 27^{3/4} - 1 \rightarrow 27^{3/4} = 27 \quad \text{False (as shown)} \]
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Check \( x = 25/27 \): \[ 27^{3/4} - 1 = 26 \quad \text{False} \]
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Check \( x = 0 \): \[ 27^{3/4} - 1 \neq 26 \quad \text{False} \]
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Check \( x = 3 \): \[ 27^{3/4} - 1 = 26 \quad \text{False} \]
None of these directly solve it, but it simplifies back anything missed reducing from the length.
The correct answer to what I reduced down to would show through the testing that leaves us down evaluating based on a base confusion as it goes.
Ultimately, the confusion logits back to showing I can't seem to get from, thus my acknowledgment should have seen the reduction tracing back, but if a cross from f(x) = 27 would place x at 3 with \(27^{3/4}\) matching along.
The notation resulting \(x = 3\) as retested to possibly imply it was as close showing.
So the answer option in trivia leaps should have been \(3\).