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If 5*125^(1/x) = 1/(5^(1/x)), then x = ?

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If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 10 May 2017, 11:41
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Question Stats:

72% (01:54) correct 28% (02:19) wrong based on 451 sessions

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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 11 May 2017, 10:06
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Top Contributor
Bunuel wrote:
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1


\(\sqrt[x]{125}\) = 125^(1/x)
= (5^3)^(1/x)
= 5^(3/x)
So, \(5*\sqrt[x]{125}\) = (5^1)[5^(3/x)] = 5^(3/x + 1)

RULE: b^(-x) = 1/(b^x)

So, \(\frac{1}{5^{\frac{1}{x}}}\) = 5^(-1/x)

Given: \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\)
We get: 5^(3/x + 1) = 5^(-1/x)
Since the bases are equal, we can conclude: 3/x + 1 = -1/x
Multiply both sides by x to get: 3 + x = -1
Solve: x = -4

Answer:

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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 10 May 2017, 13:09
Bunuel wrote:
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1



I think it should be A.

Equation can be written like 5*5^3/x=5^-1/x
On solving this we get x^2+4x=0
X=-4

Is my answer correct?
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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 10 May 2017, 23:33
1
Bunuel wrote:
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1


Ans=A

5^((3/x)+1))=5^(-1/x)
=> 4/x=-1
=>x=-4
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If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 12 May 2017, 04:27
1
2
Bunuel wrote:
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1


\(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\)
-> \(5*5^(^3^/^x^) = 5^(^-^1^/^x^)\)
-> \(5^(^1^+^3^/^x^) = 5^(^-^1^/^x^)\)
-> 1+3/x = -1/x
-> 4/x = -1
-> x= -4

Answer A.

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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 18 May 2017, 11:40
If 5*\(125^{\frac{1}{x}}\) = \(\frac{1}{5^{(1/x)}}\), then x = ?

A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1

5*\(125^{\frac{1}{x}}\) = \(\frac{1}{5^{(1/x)}}\)
\(125^{\frac{1}{x}}\) * \(5^{\frac{1}{x}}\) = \(\frac{1}{5}\)
\(5^\frac{4}{x}\) = \(5^{-1}\)
Therefore; \(\frac{4}{x}\) = -1
x = -4
Answer A...

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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 18 May 2017, 13:06
5*(125)^(1/x) = 1/5^(1/x)

=> 5^(1/x)*5*5^(3/x) = 1 = 5^0
=>1/x + 1 + 3/x = 0

Solving for x we get x = -4
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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 17 Jul 2017, 06:27
2
1
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


\(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\)


\(5 * (125)^{\frac{1}{x}} = 5^{\frac{-1}{x}}\)


\(5^1 * (5)^{\frac{3}{x}} = 5^{\frac{-1}{x}}\)


\((5)^{{1} + \frac{3}{x}} = 5^{\frac{-1}{x}}\)


\({{1} + \frac{3}{x}} = {\frac{-1}{x}}\)


\(\frac{(x + 3)}{x} = \frac{-1}{x}\)


\(x + 3 = -1\)


\(x = -4\)


Hence, Answer is A

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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 13 Jan 2019, 19:05
Bunuel wrote:
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1


Hello there!

Could someone please tell me if my process is correct? It has been a bit challenging for me to convert those kinds of terms.

Steps one and two (converting the 5^x)

Thank you in advance!

Kind regards!
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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?  [#permalink]

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New post 13 Jan 2019, 21:14
jfranciscocuencag wrote:
Bunuel wrote:
If \(5*\sqrt[x]{125}=\frac{1}{5^{\frac{1}{x}}}\), then x = ?


A. -4

B. \(\frac{-1}{\sqrt{2}}\)

C. 0

D. \(\frac{1}{\sqrt{2}}\)

E. 1


Hello there!

Could someone please tell me if my process is correct? It has been a bit challenging for me to convert those kinds of terms.

Steps one and two (converting the 5^x)

Thank you in advance!

Kind regards!


Yes, your solution is correct: \(\frac{1}{5^{\frac{1}{x}}}=5^{-\frac{1}{x}}\)
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Re: If 5*125^(1/x) = 1/(5^(1/x)), then x = ?   [#permalink] 13 Jan 2019, 21:14
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