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18 Dec 2017, 01:24
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
77% (00:38) correct
23%
(00:33)
wrong
based on 222
sessions
History
Date
Time
Result
Not Attempted Yet
23 Mar 2020, 09:38
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chetan2u
ASIDE-----------------------------
Notice the following:
k + k + k = 3k
w² + w² + w² = 3w²
xy + xy + xy = 3xy
Likewise, 3^x + 3^x + 3^x = 3(3^x)
-------------------------------------
Now let's answer the question....
Given: \(3^x+3^x+3^x=3^{21}\)
Simplify left side: \(3(3^x)=3^{21}\)
Rewrite the first \(3\) as follows: \((3^1)(3^x)=3^{21}\)
Apply the Product Law to simplify the left side: \(3^{x+1}=3^{21}\)
Now that we have the same base on both sides, we can conclude that the exponents are equal: \(x+1 = 21\)
Solve to get: \(x = 20\)
Answer: D
Cheers,
Brent
RELATED VIDEO
General Discussion
18 Dec 2017, 05:37
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chetan2u
\(3^x+3^x+3^x=3^{21}\)
i.e. \(3^x(1+1+1)=3^{21}\)
i.e. \(3^x*3=3^{21}\)
i.e. \(3^{x+1}=3^{21}\)
i.e. \(x+1 = 21\)
i.e. \(x = 20\)
ANswer: OPtion D
