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01 Oct 2025, 09:22
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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100% (01:42) correct
0% (00:00) wrong based on 1 sessions
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\(3^x × 3^y × 3^z = 729.\) If \(x, y,\) and \(z\) are three different positive integers and \(z = x + y,\) then what is the value of \(3^z\)?
04 Oct 2025, 05:55
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Official Explanation
By the law of exponents, \(3^x × 3^y × 3^z = 3^x + y + z.\) Think of this situation as 3 raised to some power must equal 729, and use trial and error on your calculator to obtain the proper exponent. You might worry that this will take too long, but exponential functions grow quickly, and in a few seconds you should obtain \(729 = 81 × 9 = 3^4 × 3^2 = 3^6.\) This means that \(x + y + z = 6.\) Observe that \(1 + 2 + 3 = 6.\) For three unalike positive integers, this is the only solution, since all other such sums will be greater than 6. Given \(z = x + y,\) then z is 3, and x and y are 1 and 2 in either order. Finally, \(3^z = 3^3 = 27.\)
Answer: 27
By the law of exponents, \(3^x × 3^y × 3^z = 3^x + y + z.\) Think of this situation as 3 raised to some power must equal 729, and use trial and error on your calculator to obtain the proper exponent. You might worry that this will take too long, but exponential functions grow quickly, and in a few seconds you should obtain \(729 = 81 × 9 = 3^4 × 3^2 = 3^6.\) This means that \(x + y + z = 6.\) Observe that \(1 + 2 + 3 = 6.\) For three unalike positive integers, this is the only solution, since all other such sums will be greater than 6. Given \(z = x + y,\) then z is 3, and x and y are 1 and 2 in either order. Finally, \(3^z = 3^3 = 27.\)
Answer: 27
