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# ((7^(1/2))^x)^2/(7^(1/2))^11)/(7^x/7^11)

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Math Expert
Joined: 02 Sep 2009
Posts: 57284

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24 Jul 2019, 02:28
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Difficulty:

35% (medium)

Question Stats:

71% (01:41) correct 29% (02:00) wrong based on 78 sessions

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$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

A. $$\sqrt{7}$$

B. 7

C. 7^2

D. 7^(11/2)

E. 7^11

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Joined: 03 Jun 2019
Posts: 1115
Location: India

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24 Jul 2019, 02:50
Bunuel wrote:
$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

A. $$\sqrt{7}$$

B. 7

C. 7^2

D. 7^(11/2)

E. 7^11

$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{(\sqrt{7})^{2x}}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{7^x}{7^{11/2}}}{\frac{7^x}{7^{11}}}$$

$$= 7^{11/2}$$

IMO D
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26 Jul 2019, 08:20
Kinshook wrote:
Bunuel wrote:
$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

A. $$\sqrt{7}$$

B. 7

C. 7^2

D. 7^(11/2)

E. 7^11

$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{(\sqrt{7})^{2x}}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{7^x}{7^{11/2}}}{\frac{7^x}{7^{11}}}$$

$$= 7^{11/2}$$

IMO D

Could you explain the last step please?
Intern
Joined: 09 Nov 2018
Posts: 2

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03 Aug 2019, 04:19
Kinshook wrote:
Bunuel wrote:
$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

A. $$\sqrt{7}$$

B. 7

C. 7^2

D. 7^(11/2)

E. 7^11

$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{(\sqrt{7})^{2x}}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{7^x}{7^{11/2}}}{\frac{7^x}{7^{11}}}$$

$$= 7^{11/2}$$

IMO D

Hey Luca1111111111111

The next step would be:
$$=\frac{7^{11}}{7^{11/2}} [m]=\frac{7^{11- 11}{2}} And hence, [m]= 7^{11/2}$$
Intern
Joined: 09 Nov 2018
Posts: 2

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03 Aug 2019, 04:25
Kinshook wrote:
Bunuel wrote:
$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

A. $$\sqrt{7}$$

B. 7

C. 7^2

D. 7^(11/2)

E. 7^11

$$\frac{\frac{((\sqrt{7})^x)^2}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{(\sqrt{7})^{2x}}{(\sqrt{7})^{11}}}{\frac{7^x}{7^{11}}}$$

$$=\frac{\frac{7^x}{7^{11/2}}}{\frac{7^x}{7^{11}}}$$

$$= 7^{11/2}$$

IMO D

Hey Luca1111111111111

The next step would be:
$$\frac{7^{11}}{7^{11/2}}$$
$$=7^{11/2}$$
Re: ((7^(1/2))^x)^2/(7^(1/2))^11)/(7^x/7^11)   [#permalink] 03 Aug 2019, 04:25
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