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505-555 (Easy)|   Exponents|      
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Bunuel
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If 5^(x + y) = 125 and 3^(x − 3y) = 1/9, then y = 24 Aug 2018, 01:20
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E
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79% (01:22) correct 21% (02:07) wrong based on 159 sessions

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If \(5^{(x+y)} = 125\) and \(3^{(x−3y)} = \frac{1}{9}\), then y =

A. −5/2
B. 1/4
C. 1/2
D. 5/2
E. 5/4
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E
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If 5^(x + y) = 125 and 3^(x − 3y) = 1/9, then y = 24 Aug 2018, 01:25
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\(5^{(x+y)}\) = 125

or \(5^{(x+y)}\) = \(5^3\)

x+y = 3......(1)

Also,

\(3^{(x−3y)}\) = \(\frac{1}{9}\) = \(\frac{1}{3^2}\) or \(3^{-2}\)

x-3y = -2.......(2)

Solving 1 and 2 gives 5 = 4y or Y = \(\frac{5}{4}\).
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If 5^(x + y) = 125 and 3^(x − 3y) = 1/9, then y = 24 Aug 2018, 01:30
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Bunuel
If \(5^{(x+y)} = 125\) and \(3^{(x−3y)} = \frac{1}{9}\), then y =

A. −5/2
B. 1/4
C. 1/2
D. 5/2
E. 5/4
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\(5^{(x+y)} = 125\)
Or, \(5^{(x+y)} = 5^3\)
Or, x+y=3-----(1)

\(3^{(x−3y)} = \frac{1}{9}\)=\(3^{-2}\)
Or, x-3y=-2--(2)
From(1) and (2), we have,

4y=5
Or, y=5/4

Ans. (E)
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If 5^(x + y) = 125 and 3^(x − 3y) = 1/9, then y = 24 Aug 2018, 01:54
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Bunuel
If \(5^{(x+y)} = 125\) and \(3^{(x−3y)} = \frac{1}{9}\), then y =

A. −5/2
B. 1/4
C. 1/2
D. 5/2
E. 5/4
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Given ,

\(5^{(x + y)} = 5^3\)

x + y = 3

and

\(3^{(x - 3y)}= 3^{(-2)}\)

x - 3y = -2

We are asked to find out the value of y.

x + y - (x - 3y) = 3 - (-2)

x + y - x + 3y = 5

4y = 5

y =\(\frac{5}{4}\)


The best answer is E.
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If 5^(x + y) = 125 and 3^(x − 3y) = 1/9, then y = 08 May 2020, 02:04
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Bunuel
If \(5^{(x+y)} = 125\) and \(3^{(x−3y)} = \frac{1}{9}\), then y =

A. −5/2
B. 1/4
C. 1/2
D. 5/2
E. 5/4
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\(5^{(x+y)} = 125\)

So, \(x + y = 3\)

And, \(3^{(x−3y)} = \frac{1}{9}\)

Or, \(3^{(x−3y)} = \frac{1}{3^2}\)

Or, \(3^{(x−3y)} =3^{-2}\)

So, \(x - 3y = -2\)


Now solve -

\(-x -y = -3\) + \(x - 3y = -2\)

Or, \(-4y = -5\)

So, \(y = \frac{5}{4}\), Answer must be (E)
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If 5^(x + y) = 125 and 3^(x − 3y) = 1/9, then y = 08 May 2020, 03:34
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\(5^{x+y}\) = 125. Since 125 =\( 5^3\),
\(5^{x+y}\) =\( 5^3\). Let’s call this equation (1)

\(3^{x-3y}\) = \(\frac{1}{9}\). Since 9 = \(3^2\), 1/9 = \(\frac{1}{3^2}\) = \(3^{-2}\).

Here we have used the rule that \(\frac{1}{x^{n}}\) = \(x^{-n}\)

Therefore, \(3^{x-3y}\) = \(3^{-2}\). Let’s call this equation (2)

In both equations (1) and (2), since we have equal numbers with the same bases, the exponents have to be equal. Therefore,
x+y = 3
x – 3y = -2.

Solving, we have, y = \(\frac{5}{4}\).

The correct answer option is E.
Hope that helps!
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