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Bunuel
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Bunuel
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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Bunuel
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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Bunuel
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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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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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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Bunuel
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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Bunuel
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jfranciscocuencag
Bunuel
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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why x = 0 is not considered in this case ?
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shashank07!
why x = 0 is not considered in this case ?

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

How would \(x = 0\) satisfy the equation? The 0th root of a number is not defined (so \(\sqrt[x]{125}\) would be undefined), and division by 0 is also undefined. Therefore, \(5^{^{\frac{1}{x}}}\) would be undefined as well.
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