For real numbers a and b, is it true that a = b?

For real numbers a and b, is it true that a = b?

(1) a^4 = b^4 --> |a| = |b| --> a = b or a = -b. Nor sufficient.

(2) a^5 = b^5 --> a = b. Sufficient.

Answer: B.
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Real numbers are any number ....

By using both statements we can say c can be .. the answer

Bcz in B , a^5 = b^5

A can be 2^ 4/5. And b can be 4^ 2/5

But both ans will be same .


Thus we need both 1 and 2 to solve this question

C
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Yes it is B. However I don't understand why.

1) Since a and b are both raised to the fourth power, the answers will be positive. But, a could equal negative b. For example, a = 2, b = -2.
2) Since both are raised to the 5th power, a = b. For example, a = b = 2 OR a = b = -2.

Hope this is correct!

Nahh! not true. For example, by plugging in numbers.

St1: a^4 = b^4 --> if we plugin a = 2 & b =2 --> 2^4 = 2^4 --> 16 = 16 so right whenever a^4 is equal to b^4 then a will be equal to b .... but if one of them is negative number while other is positive, they yield the same result but in that case a is not equal to b --> a = 2, b = -2 --> 2^4 = (-2)^4 --> 16 = 16 hence a^4 = b^4 but a is not equal to b. Same result can be yielded for a^4 = b^4 whether +ve & -ve or +ve & +ve or -ve & -ve same numbers are plugged in which case either a can be equal to b or cannot be.

St1 alone Insufficient


St2: both a & b need to be same +ve numbers for a^5 to be equal to b^5 or both a & b be -ve numbers for a^5 to be equal to b^5. If one is +ve whereas the other is -ve then a^5 won't be equal to b^5 that's why for a^5 = b^5 either both a & b have to be +ve or -ve.

a^5 = b^5 --> plugin both +ve yet same number by having a = 2 , b = -2 --> 2^5 = 2^5 --> 32 = 32 hence a^5 = b^5.

And, now --> a^5 = b^5 --> plugin both -ve yet same numbers by having a = -2 , b = -2 --> (-2)^5 = (-2)^5 --> -32 = -32 hence a^5 = b^5.

What if a = 2 & b = -2 in which case they are not same then for a^5 = b^5 --> 2^5 = (-2)^5 --> 32 = -32 which can't be true hence a^5 can't be equal to b^5. Hence to get a^5 = b^5 we must have a = b.

St2 alone Sufficient

So, answer should be B

It is an easy question and I went into the detail so that it may be easy for others, who may be new to the forum, to understand otherwise answer provided by Bunuel was enough.

(1) a^4 = b^4
Four possible conditions:-
1) a=b=0
2) a=b
3) a= -b (even powers give +ve numbers)
4) -a=b
Not sufficient

(2) a^5 = b^5
three possible conditions:-

a=b=0
a=b
-a=-b
Sufficient

B is the answer
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The first thing to notice is that this is a yes/no question. We don't need to know exact values for a and b. We just need enough information to prove that a and be are the same.

Next, a very important difference between odd powers and even powers:

For every positive number, there are two numbers that can be raised to an EVEN power to yield that number. For example both 2 and -2 can can be squared to get 4. Similarly, both the positive 6th root of 732 and the negative 6th root of 732 can be raised to the 6th power to get 732.

However, for every number (both pos and neg), there is only one number that can be raised to an odd power to yield that number. For example, the only number we can raise to the 3rd power to get 27 is 3.

It's worth noting what this means for algebra. You CAN take an odd root of both sides of an equation. But you cannot reliably take an even root of both sides of an equation.

So the statements:

Statement 1) so both a and be lead to the same number when they are raised to an EVEN power. But a and b aren't necessarily the same number since one might be pos and the other might be neg. For example a could be 2 and b could be -2, but they also might both be 2. Insufficient.

Statement 2) both a and b yield the same number when they are raised to an ODD power. The only way this is possible is for a and b to be the same number. Sufficient.

Bunuel,

Could you please let me understand why in statement 2 we assume only the case when a=b like(+-1, +-1/2 and so on), what if a = 5th√(32^5), which gives us 32 and b = 2^5 which gives us again 32, thus a^5=b^5, while a=√(32^5) and b=2?

Thanks a lot

\(a=\sqrt[5]{32^5}=32\)
\(b=2^5=32\)

Doesn't a = b?
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As real numbers could be both positive and negative,

(1) will be true even if the sign of both the numbers is different as it has an even power - INSUFFICIENT

(2) will be true only if the numbers have the same sign - SUFFICIENT

Answer is B

Bunuel,

yes, that is true. Sorry, don't know why I made this simple question look overcomplecated for me. I was thinking like what if a=√(32^5) (which is 32^5/2), and when we have a^5 we actually have (√(32^5))^5 (which is 32^25/2) here is my mistake, I thought (√(32^5))^5 = just 32 , which can be rewritten as 2^5 (then I have a =√(32^5) and b=2, whcih is not equal to a)

Thank you!

The catch is real numbers a and b.

(1) (-2) ^ 4 = (2) ^4. The numbers a and b are not necessarily required to be equal in this case. Therefore the statement is not sufficient.

(2) Real numbers a and b have to be equal for their odd powered values to be equal. 2^5 is only equal to 2^5.

B
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(1) a^4 = b^4 --> |a| = |b| --> a = b or a = -b. Nor sufficient.

(2) a^5 = b^5 --> a = b. Sufficient.

statement 1, using absolute value, |a|=|b| same judgment why not/why not test statement 2???

I need an easy solution, i can not understand.

(1) a=b or a=-b . Not Sufficient
(2) a=b . Sufficient

Hence, OA is (B).

Bunuel, this was asked in GMAT Prep Mock 5. Could you please tag it?

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Done. Thank you.
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