Question
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Answer
When \( x = 2 \), the square equals 1.
Solution
We start with the equation
\[
8^{2x-3} = 8^1.
\]
Since the bases are the same, we can set the exponents equal:
\[
2x - 3 = 1.
\]
Now, we solve for \(x\):
1. Add 3 to both sides:
\[
2x = 1 + 3 = 4.
\]
2. Divide both sides by 2:
\[
x = \frac{4}{2} = 2.
\]
Thus, the solution is \(x=2\). Given the problem statement, we can complete the statement by noting that the exponent equals 1, so
\[
\square = 1 \quad \text{when} \quad x=2.
\]
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Simplify this solution Bonus Knowledge
To solve the equation \( 8^{2x-3} = 8^{1} \), we can utilize the property of exponents that states if the bases are the same, we can set the exponents equal to each other. This means that we can directly equate the exponents: \[ 2x - 3 = 1 \] Now, we can solve for \( x \): 1. Add 3 to both sides: \[ 2x = 1 + 3 \] \[ 2x = 4 \] 2. Divide both sides by 2: \[ x = \frac{4}{2} = 2 \] Thus, substituting for \( \square \), the answer is \( \boxed{2} \). To double-check, we can substitute \( x \) back into the original equation: \[ 8^{2(2) - 3} = 8^{1} \] \[ 8^{4 - 3} = 8^{1} \] \[ 8^{1} = 8^{1} \] The original equation holds true!
