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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=
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13 Jul 2022, 07:22
13 Jul 2022, 07:22
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Question Stats:
70% (03:27) correct
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If \(j ≠ |k|\), \(jk ≠ 0\), and \(j^{2x}=k^{2y}=jk\), then \(x =\)
A. \(\frac{4y}{2y-1}\)
B. \(\frac{2y}{2y-1}\)
C. \(\frac{y}{2y-1}\)
D. \(\frac{y^2}{2y-1}\)
E. \(\frac{2y-1}{y^2}\)
A. \(\frac{4y}{2y-1}\)
B. \(\frac{2y}{2y-1}\)
C. \(\frac{y}{2y-1}\)
D. \(\frac{y^2}{2y-1}\)
E. \(\frac{2y-1}{y^2}\)
Re: If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=
[#permalink]
13 Jul 2022, 16:23
13 Jul 2022, 16:23
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My answer is B.
Stage 1
j^(2x) = jk
(j^(2x))/j = k
j^(2x-1) = k --> Equation 1
Similarly for k...
k^(2y) = jk
k^(2y-1) = j --> Equation 2
Substitute k from Equation 1 into Equation 2.
(j^(2x-1))^(2y-1) = j
j^(4xy - 2x - 2y + 1) = j
This means the exponent on the left hand side must equal 1 because the right hand side can be rewritten as j = j^1.
4xy - 2x - 2y + 1 = 1
2x*(2y - 1) = 2y
x*(2y-1) = y
x = y/(2y-1) --> Answer B
Stage 1
j^(2x) = jk
(j^(2x))/j = k
j^(2x-1) = k --> Equation 1
Similarly for k...
k^(2y) = jk
k^(2y-1) = j --> Equation 2
Substitute k from Equation 1 into Equation 2.
(j^(2x-1))^(2y-1) = j
j^(4xy - 2x - 2y + 1) = j
This means the exponent on the left hand side must equal 1 because the right hand side can be rewritten as j = j^1.
4xy - 2x - 2y + 1 = 1
2x*(2y - 1) = 2y
x*(2y-1) = y
x = y/(2y-1) --> Answer B
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=
[#permalink]
14 Jul 2022, 06:33
14 Jul 2022, 06:33
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GreenlightTestPrep wrote:
If \(j ≠ |k|\), \(jk ≠ 0\), and \(j^{2x}=k^{2y}=jk\), then \(x =\)
A. \(\frac{4y}{2y-1}\)
B. \(\frac{2y}{2y-1}\)
C. \(\frac{y}{2y-1}\)
D. \(\frac{y^2}{2y-1}\)
E. \(\frac{2y-1}{y^2}\)
A. \(\frac{4y}{2y-1}\)
B. \(\frac{2y}{2y-1}\)
C. \(\frac{y}{2y-1}\)
D. \(\frac{y^2}{2y-1}\)
E. \(\frac{2y-1}{y^2}\)
Take: \(k^{2y}=jk\)
Divide both sides of the equation by \(k\) to get: \(\frac{k^{2y}}{k}=j\)
Simplify the left side: \(k^{2y-1}=j\)
Now take: \(j^{2x}=k^{2y}\)
Replace \(j\) with \(k^{2y-1}\) to get: \((k^{2y-1})^{2x}=k^{2y}\)
Apply the power of a power law to the left side: \(k^{(2x)(2y-1)}=k^{2y}\)
Since the bases are equal, we can write: \((2x)(2y-1) = 2y\)
Divide both sides of the equation by \(2y-1\) to get: \(2x = \frac{2y}{2y-1}\)
Divide both sides of the equation by \(2\) to get: \(x = \frac{y}{2y-1}\)
Answer: C
