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# For non-negative x and y, is (0.8)^x+(0.6)^x<y ?

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Math Expert
Joined: 02 Sep 2009
Posts: 58402
For non-negative x and y, is (0.8)^x+(0.6)^x<y ?  [#permalink]

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11 Apr 2017, 01:29
00:00

Difficulty:

45% (medium)

Question Stats:

63% (01:38) correct 37% (01:39) wrong based on 150 sessions

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For non-negative x and y, is $$(0.8)^x+(0.6)^x<y$$ ?

(1) y ≥ 1

(2) x ≥ 3

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Manager
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Joined: 25 Jan 2016
Posts: 127
Location: India
Concentration: Finance, Entrepreneurship
GPA: 4
WE: Web Development (Computer Software)
Re: For non-negative x and y, is (0.8)^x+(0.6)^x<y ?  [#permalink]

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11 Apr 2017, 08:16
1

for x=1 and x=2 value is 1 so combining we can solve the question
Manager
Joined: 03 Sep 2018
Posts: 177
For non-negative x and y, is (0.8)^x+(0.6)^x<y ?  [#permalink]

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16 Jan 2019, 06:51
$$\frac{4}{5}^x + \frac{3}{5}^x < 1$$

Now, when x<0 this might not be true, hence we need both statements
Intern
Joined: 05 Mar 2019
Posts: 16
Re: For non-negative x and y, is (0.8)^x+(0.6)^x<y ?  [#permalink]

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27 Mar 2019, 02:11
1
ghnlrug wrote:
$$\frac{4}{5}^x + \frac{3}{5}^x < 1$$

Now, when x<0 this might not be true, hence we need both statements

X is non-negative per question statement
Re: For non-negative x and y, is (0.8)^x+(0.6)^x<y ?   [#permalink] 27 Mar 2019, 02:11
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