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Sub 505 (Easy)|   Exponents|                                       
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Walkabout
If mˉ¹ = -1/3 then mˉ² is equal to

(A) -9
(B) -3
(C) -1/9
(D) 1/9
(E) 9

Some good solutions here.
Here's one more approach:

CONCEPT: b^(-n) = 1/(b^n)

Given: mˉ¹ = -1/3
Rewrite as: 1/m = 1/(-3)
We can see that m = -3

So, mˉ² = 1/(m²) = 1/(-3)² = 1/9

Answer:
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Walkabout
If m^(-1) = -1/3 then m^(-2) is equal to

(A) -9
(B) -3
(C) -1/9
(D) 1/9
(E) 9

Since we know that m^-1 = -1/3 and m^-2 = (m^-1)^2, then m^-2 = (-1/3)^2 = 1/9.

The answer is D.
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M^-1=-1/3=> M=-3
hence M^-2=1/9
Smash D
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m^(-1) = -(1/3)
Squaring both sides
m^(-2) = 1/9

Ans. D
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\(m^-1 = \frac{-1}{3}\)

\(\frac{1}{m} = \frac{-1}{3}\)

By cross multiplication

\(-m = 3\)

Squaring both sides

\(m^2 = 9\)

So \(m^-2 = \frac{1}{9}\)
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m^(−1)=−1/3
now,squaring both sides
[m^(−1)]^2=[−1/3]^2
m^(−2)=1/9.

correct answer D
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Walkabout
If \(m^{(-1)} = -\frac{1}{3}\) then \(m^{(-2)}\) is equal to

(A) -9
(B) -3
(C) -1/9
(D) 1/9
(E) 9

Pretty simple and straight forward, doesn't really require any calculations.
Base raised to negative exponent = \(\frac{1}{m} = -\frac{1}{3}\) Thus \(\frac{1}{m^2} = (-\frac{1}{3})^2 = \frac{1}{9}\)
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Walkabout
If \(m^{(-1)} = -\frac{1}{3}\) then \(m^{(-2)}\) is equal to

(A) -9
(B) -3
(C) -1/9
(D) 1/9
(E) 9





Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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