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06 May 2019, 10:15
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:35) correct
60%
(02:42)
wrong
based on 67
sessions
History
Date
Time
Result
Not Attempted Yet
How many real numbers x satisfy the equation 3^(2x+2) − 3^(x+3) − \(3^x\) + 3 = 0?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
07 May 2019, 02:31
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Rubina11
\(3^(2x+2) − 3^(x+3) − 3^x + 3 = 0\)
Or, \(3^{2x}*3^2-3^x*3^3-3^x+3=0\)
Or, \((3^x)^2*3^2-3^x*3^3-3^x+3=0\)
Suppose \(3^x=t\), substituting, we have
\(t^2*3^2-t*3^3-t+3=0\)
Or, \(9t^2-28t+3=0\) Upon factorization of this quadratic eq, we have
\(9t^2-27t-t+3=0\)
Or, 9t(t-3)-1(t-3)=0
Or, (t-3)(9t-1)
Or. t=3 ; \(t=\frac{1}{9}\)
1) \(3^x=3\) or x=1
2) \(3^x=3^{-2}\) or x=-2
Ans. (C)
07 May 2019, 03:11
Kudos
Bookmarks
Rubina11
\(3^{2x+2} − 3^{x+3} − 3^x+ 3 = 0\)
\(9*3^{2x} - 27*3^x - 3^x + 3 = 0\)
\(9*3^{2x} - 28*3^x + 3 = 0\)
Let \(3^x = a\)
\(9a^2 - 28a + 3 = 0\)
\((9a - 1)*(a - 3) = 0\)
a = 1/9, 3
\(3^x = 1/9 = 1/3^2 = 3^{-2}\)
x = -2
\(3^x = 3\)
x = 1
Two real values of x satisfy the equation.
Answer (C)
