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Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of

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Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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New post 09 Jan 2019, 07:54
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65% (01:32) correct 35% (02:06) wrong based on 31 sessions

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Given that \(10^\frac{48}{100}\) = x, \(10^\frac{7}{10}\) = y and\(x^z\) = \(y^2\), then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36

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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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New post 09 Jan 2019, 08:10
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fitzpratik wrote:
Given that \(10^\frac{48}{100}\) = x, \(10^\frac{7}{10}\) = y and\(x^z\) = \(y^2\), then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36


GIVEN: \(10^\frac{48}{100}\) = x
Rewrite as: \(10^{0.48}\) = x

GIVEN: \(10^\frac{7}{10}\) = y
Rewrite as: \(10^{0.7}\) = y


GIVEN: \(x^z\) = \(y^2\)

Replace with above values to get: \((10^{0.48})^z\) = \((10^{0.7})^2\)

Apply Power of a Power Law to get: \(10^{0.48z}=10^{1.4}\)

This means: \(0.48z = 1.4\)

Divide both sides by 0.48 to get: \(z = \frac{1.4}{0.48}\)

NOTE: Since \(\frac{1.4}{0.5}=2.8\), we know that \(\frac{1.4}{0.48}\) is a little bit greater than 2.8

Check the answer choices . . .
Answer: C

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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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New post 09 Jan 2019, 08:29
fitzpratik wrote:
Given that \(10^\frac{48}{100}\) = x, \(10^\frac{7}{10}\) = y and\(x^z\) = \(y^2\), then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36


I just ball parked in this question.
\(10^\frac{48}{100}\) = x, \(10^\frac{7}{10}\) = y

\(x^z\) = \(y^2\)

\(10^\frac{48}{100}\) = x
\(10^\frac{1}{2}\) = x

\(x^z\) = \(y^2\)

\(10^\frac{z}{2}\) = \(10^\frac{14}{10}\)
\(\frac{z}{2}\)= \(\frac{7}{5}\)

z = 2.85

Answer C
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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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New post 09 Jan 2019, 09:00
fitzpratik wrote:
Given that \(10^\frac{48}{100}\) = x, \(10^\frac{7}{10}\) = y and\(x^z\) = \(y^2\), then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36



\(y = 10^{\frac{7}{10}}\)

\(y^2 = 10^{\frac{7}{10} * 2} = 10^{\frac{7}{5}}\)

\(x = 10^{\frac{48}{100}} = 10^{\frac{12}{25}}\)

\(x^z = 10^{\frac{12}{25}}^z = 10^{\frac{7}{5}} = y^2\)

So \((\frac{12}{25})z = \frac{7}{5}\)

\(z = \frac{35}{12} = 2.9\)

Answer (C)
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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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New post 09 Jan 2019, 10:32
We get

10^(48z/100) =10^14/10

Equating exponents (since we are in base 10) gives 48z/100 =14/10
then 480z=1400
and z=1400/480. 480*3=1440, therefore the answer is a little less than 3. C
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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of   [#permalink] 09 Jan 2019, 10:32
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