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\frac{(([square_root]7[/square_root])^x)
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17 Apr 2019, 01:49
17 Apr 2019, 01:49
Expert Reply
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Question Stats:
60% (01:02) correct
39% (00:58) wrong based on 61 sessions
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Quantity A |
Quantity B |
\(\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}\) |
\(\frac{7^x}{7^{11}}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: \frac{(([square_root]7[/square_root])^x)
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23 Apr 2019, 08:23
23 Apr 2019, 08:23
Any solutions to this please?
Appreciated!
Appreciated!
Re: \frac{(([square_root]7[/square_root])^x)
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23 Apr 2019, 12:00
23 Apr 2019, 12:00
1
Expert Reply
This is a tough one. However, you should visualize the solution up-front before to get into unuseful calculations.
QA \((\sqrt{7}) ^ {2x}\) (multiply the exponents)
\(\frac{(\sqrt{7})^{2x}}{(\sqrt{7})^{11}}\)
For the root's properties, the root of 7 becomes \(7^{\frac{2x}{2}}\) and the denominator is \(7^{\frac{11}{2}}\).
From this \(7^x\) and \(7^{5.5}\)
\(\frac{7^x}{7^{5.5}}\) VS \(\frac{7^x}{7^{11}}\)
Everything boils down to compare the denominator of the two fractions \(7^{11} > 7^{5.5}\) as a number.
However, we must consider the fraction \(\frac{1}{7^{5.5}}\) which is > of \(\frac{1}{7^{11}}\)
Therefore, A must be bigger
Hope this helps.
Regards
QA \((\sqrt{7}) ^ {2x}\) (multiply the exponents)
\(\frac{(\sqrt{7})^{2x}}{(\sqrt{7})^{11}}\)
For the root's properties, the root of 7 becomes \(7^{\frac{2x}{2}}\) and the denominator is \(7^{\frac{11}{2}}\).
From this \(7^x\) and \(7^{5.5}\)
\(\frac{7^x}{7^{5.5}}\) VS \(\frac{7^x}{7^{11}}\)
Everything boils down to compare the denominator of the two fractions \(7^{11} > 7^{5.5}\) as a number.
However, we must consider the fraction \(\frac{1}{7^{5.5}}\) which is > of \(\frac{1}{7^{11}}\)
Therefore, A must be bigger
Hope this helps.
Regards
