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GMATMadeeasy
Question : \(a^b^c\)

How one will interpret this ?\((a^b)^c\) or \((a)^(b^c)\)

Second :
\((81)^1/4\) ; It is equivalent to 3 and or -3 in GMAT ? I am aware that \(\sqrt{(x^2)}\) is always positive value.

First question: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), for instance \(3^{3^3}=3^{(3^3)}=3^{27}\).

Second question: \(\sqrt[_{even}]{x}\geq0\) - Even roots have only a positive value on the GMAT. (well if x=0 then it will obviously be 0). So \(\sqrt[4]{81}=3\), not +3 and -3.

Also general rule: \(\sqrt{x^2}=|x|\).

When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) can not be negative but \(x\) can.

\(y\)can not be negative as \(y=\sqrt{some expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.

That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation \(x^2 = 16\) has TWO solutions, \(x=+4\)and \(x=-4.\)

On the other hand odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{64}=4\) and \(\sqrt[3]{-27}=-3\).

Hope it's clear.
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Let's try to solve this based on above.

What is the value of x ?

1 \sqrt{x^4} = 9
2 The average of x^2,6x and 3 is -2


P.S. First atement is square root of (x^4)
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Let's try to solve this based on above.

What is the value of x ?

1 \sqrt{x^4} = 9
2 The average of x^2,6x and 3 is -2


P.S. First atement is square root of (x^4)

(1) I would think this gives x=+3 or x=-3 as possible solutions, so not sufficient.
(2) (x^2 + 6x + 3)/3 =-2
x^2 + 6x +3 = -6
x^2 + 6x + 9 = 0
x=-3

So (2) alone, B?
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B;

agree with andrewcs... A doesnt give unique value, B does...
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