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If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?

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If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?  [#permalink]

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16 Feb 2017, 02:23
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5% (low)

Question Stats:

85% (01:09) correct 15% (01:25) wrong based on 175 sessions

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If $$\sqrt{10x}=10\sqrt{10}$$, and $$\frac{10^{(-5)}}{10^{(-y)}}=x$$, what is y?

A. -2
B. 0
C. 5
D. 6
E. 7

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Re: If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?  [#permalink]

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16 Feb 2017, 02:39
1
Bunuel wrote:
If $$\sqrt{10x}=10\sqrt{10}$$, and $$\frac{10^{(-5)}}{10^{(-y)}}=x$$, what is y?

A. -2
B. 0
C. 5
D. 6
E. 7

$$\sqrt{10x}=10\sqrt{10}$$
10x =1000 =>x=100 =>x =$$10^2$$

$$\frac{10^{(-5)}}{10^{(-y)}}=x$$
$$10^{(-5+y)}=x=10^2$$
y-5=2
y=7

Hence Option E is correct.
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Re: If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?  [#permalink]

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21 Feb 2017, 09:29
1
Bunuel wrote:
If $$\sqrt{10x}=10\sqrt{10}$$, and $$\frac{10^{(-5)}}{10^{(-y)}}=x$$, what is y?

A. -2
B. 0
C. 5
D. 6
E. 7

Let’s first determine x:

√(10x) = 10√10

(√10)(√x) = 10√10

√x = 10

x = 100

Now let’s substitute x into the second equation.

10^-5/10^-y = 100

10^(-5 + y) = 10^2

-5 + y = 2

y = 7

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Re: If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?  [#permalink]

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03 Apr 2017, 23:32
Squaring both the sides, we get:

10x=100*10
x=100=10^2

10^y/10^5= x
x= 10^y-5

10^2= 10^y-5

y-5=2
y= 7
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Re: If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?  [#permalink]

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09 Jan 2019, 03:25
ScottTargetTestPrep wrote:
Bunuel wrote:
If $$\sqrt{10x}=10\sqrt{10}$$, and $$\frac{10^{(-5)}}{10^{(-y)}}=x$$, what is y?

√x = 10

x = 100

ScottTargetTestPrep why do we not have to check for extraneous solutions? would sqrt(x)=10 mean ±x=100?
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Re: If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y?  [#permalink]

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10 Jan 2019, 09:49
1
ghnlrug wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
If $$\sqrt{10x}=10\sqrt{10}$$, and $$\frac{10^{(-5)}}{10^{(-y)}}=x$$, what is y?

√x = 10

x = 100

ScottTargetTestPrep why do we not have to check for extraneous solutions? would sqrt(x)=10 mean ±x=100?

Sqrt(x) = 10 always implies x = 100, x = -100 is not a solution here (simply because sqrt(-100) is not 10; it is undefined when you are working with real numbers). The square root function is only defined for positive real numbers; therefore if you are taking the square root of some number and fining 10, the number you had to begin with can only be 100.

One of the situations where you would have extraneous solutions is an equation like x^2 = 100. If this was the case, then both x = 10 and x = -10 would be legitimate solutions (because squaring both will give you 100). On the other hand, sqrt(100) is always 10; in fact, sqrt(a) can be defined as the positive root of the equation x^2 = a.
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Re: If (10x)^(1/2)=10*10^(1/2), and 10^(-5)/10^(-y)=x, what is y? &nbs [#permalink] 10 Jan 2019, 09:49
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