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24 Aug 2021, 08:00
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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:
55%
(hard)
Question Stats:
65% (02:20) correct
35%
(02:36)
wrong
based on 292
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If \(3^{(2x+1)} + 4(3^x) – 20 = 0\), then what is approximate value of x ?
A. 1/6
B. 1/5
C. 1/3
D. 2/3
E. 9/10
M37-26
A. 1/6
B. 1/5
C. 1/3
D. 2/3
E. 9/10
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M37-26
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22 Nov 2021, 06:27
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Official Solution:
If \(3^{(2x+1)} + 4(3^x) – 20 = 0\), then what is approximate value of \(x\) ?
A. \(\frac{1}{6}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(\frac{9}{10}\)
\(3^{(2x+1)} + 4(3^x) – 20 = 0\);
\(3*3^{(2x)} + 4(3^x) – 20 = 0\);
\(3*(3^x)^2 + 4(3^x) – 20 = 0\);
Denote \(3^x\) as \(k\): \(3k^2 + 4k– 20 = 0\);
Solve for \(k\): \(k=2\) or \(k=-\frac{10}{2}\) (discard this root because \(3^x\) is positive and cannot equal to a negative number).
We get \(3^x=2\)
Now, \(3^2 \approx 2^3\), so \(3^{\frac{2}{3}} \approx 2\), thus \(x\approx {\frac{2}{3}}\).
Answer: D
If \(3^{(2x+1)} + 4(3^x) – 20 = 0\), then what is approximate value of \(x\) ?
A. \(\frac{1}{6}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(\frac{9}{10}\)
\(3^{(2x+1)} + 4(3^x) – 20 = 0\);
\(3*3^{(2x)} + 4(3^x) – 20 = 0\);
\(3*(3^x)^2 + 4(3^x) – 20 = 0\);
Denote \(3^x\) as \(k\): \(3k^2 + 4k– 20 = 0\);
Solve for \(k\): \(k=2\) or \(k=-\frac{10}{2}\) (discard this root because \(3^x\) is positive and cannot equal to a negative number).
We get \(3^x=2\)
Now, \(3^2 \approx 2^3\), so \(3^{\frac{2}{3}} \approx 2\), thus \(x\approx {\frac{2}{3}}\).
Answer: D
General Discussion
24 Aug 2021, 08:16
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My answer is D).
Assume 3^x = A, then the equation becomes: 3A^2+4A-20=0 -> (A-2)(3A+10)=0 -> A=2 (3A+10>0).
So 3^x = 2. Looking at the choices: A)-C) can be quickly eliminated because say if x = 1/3, then 3^(1/3) = 2, 3 = 2^3, clearly not going to work.
E) can also be eliminated because by the same logic, 3^9 = 2^10. We know that 3^x will climb much faster than 2^x when x becomes larger.
D) works because if 3^(2/3) = 2, then 3^2 = 9 = 2^3 = 8, close enough. So D).
Assume 3^x = A, then the equation becomes: 3A^2+4A-20=0 -> (A-2)(3A+10)=0 -> A=2 (3A+10>0).
So 3^x = 2. Looking at the choices: A)-C) can be quickly eliminated because say if x = 1/3, then 3^(1/3) = 2, 3 = 2^3, clearly not going to work.
E) can also be eliminated because by the same logic, 3^9 = 2^10. We know that 3^x will climb much faster than 2^x when x becomes larger.
D) works because if 3^(2/3) = 2, then 3^2 = 9 = 2^3 = 8, close enough. So D).
