If a and b are different nonzero integers, what is the value of b? (1

Got tricked initially (did the same mistake), good question !! (need to take note that a,b are different non-zero integers)

Q : If a and b are different and nonzero integers , what is the value of b?

make sure to use all the information provided in question stem, question stem tells us that

(i) \(a\neq{0}\) or \(b\neq{0}\)
(ii) \(a\neq{b}\)

now to the statements
stmt 1: \(a^{b}\) = ab

(a,b) = (1,1), (2,2) , (3,1) , (4,1)

1^1 = 1*1 --> ignore this a,b are different non-zero
2^2 = 2*2 --> ignore this a,b are different non-zero
3^1 = 3*1
4^1 = 4*1...

b is always 1---> b =1 got one solution , so Sufficient

stmt 2: \(a^{b} - a^{b-1}\)= 2

Simplify : \(a^{b}\) (1-\(\frac{1}{a}\)) =2

this equation works for values as above (a,b) = (2,2) , (3,1) (refer below with inserted values)

\(2^{2}\) (1-\(\frac{1}{2}\)) => 4 (\(\frac{1}{2}\)) = 2

(2,2) this works But, ignore this a,b are different non-zero

\(3^{1}\) (1-\(\frac{1}{3}\)) => 3 (\(\frac{2}{3}\)) = 2

(a,b) = (3,1) works

[Edited] got one more solution

\((-1)^{2}\) (1-\(\frac{1}{(-1)}\)) => 1 (1+1) = 2
(a,b) = (-1,2) did works ---> b got two solution , Not Sufficient

so both stmt 1 & stmt2 are Sufficient:

Ans : A

What about a=2 and b=2

2^2 = 4

viksingh15
it is tricky, make sure to use all the information provided in question stem.I understand the confusion.

Q : If a and b are different and nonzero integers , what is the value of b?

it tells us that

1> \(a\neq{0}\) or \(b\neq{0}\)

2> \(a\neq{b}\)

MANHATTAN GMAT OFFICIAL SOLUTION:

(1) SUFFICIENT: If b=1, then a can be any number whatsoever, because a^1 and a(1) will always be the same number. Therefore the issue is to determine whether b can be any number other than 1.

The given equation can be rephrased. Divide both sides by a, giving a^{b-1}=b.

First, consider the case in which b is positive.

If b=2, then this equation reduces to a^1=2, so that a=2; however, this case can be disregarded, because a and b are not allowed to be the same.

If b=3 or more, the equation is a^2=3, a^3=4, a^4=5, etc. None of these equations have integer solutions.

What if b is negative?

If b=-1, then a^{-2}=-1, which is impossible! Something raised to an even power cannot be negative.

If b=-2 or less, then we have a^{-3}=-2, a^{-4}=-3, etc. These equations are all impossible. If the power is even, then the expression can’t be negative. If the power is odd, the resulting value of a will be a fraction.

This rules out every case except b=1. The statement is sufficient.

(2) NOT SUFFICIENT: To find cases that simplify this equation, it is helpful to factor it. Pull an a^{b-1} out of both terms on the right side: a^{b-1}(a-1)=2. (If you’re not sure why what’s the right factorization, write out the steps yourself; that’ll be a great test of your knowledge of exponent rules).

Now test some cases:

If a=3 and b=1, then the expression is 3^0(3-1)=1(2)=2.

If a=1 and b=-2, then the expression is -1^{-3}(-1-1)=(-1)(-2)=2.

There are at least two possibilities for b, so this statement can’t be sufficient. It turns out that if a=-1 and b= any even number, you’ll get 2 as the answer (not sure why? test some even numbers to see whether you can figure out the rule). Therefore, b can be 1 or any even number.

The correct answer is A.

a≠b and a ≠0

1)
ab=ab
a(a^(b-1) -1 ) = 0

either a =0 or ab-1 =1
if and b are intergers only way possible for this is for b = 1
2)
a^b−a^(b−1)=2
a^b ( 1 – 1/a) = 2
a^( b-1) * ( a- 1) = 2
( a = 3 and b = 1 )and (a =2 and b = 2) both will satisfy this equation.

Answer A

I see that many people have explained it correctly, but also note that though they have mentioned a=-1, b=positive even number works in 2, but they have not explained why it works, I have also explained why a=-1 works in point 2.

1) \(a^b = ab\)
\(a^{b-1} = b\) (as \(a\neq 0\))
Lets see what range of value b can take
\(b\leq0\): b cant be \(\leq0\) because then \(a^{b-1}\) will be fraction
\(b=1\): possible then LHS=1 as anything raised to the power 0 is 1. RHS=b=1
\(b>1\): b cant be >1 because then \(a^{b-1}\) will be larger than b. (Because a>1, if a=1, then b also has to be 1 which is not allowed as per question as the two integers are different)
As b can take only one value i.e. 1; therefore, it is sufficient.

2)\(a^b - a^{b - 1} = 2\)
If a=-1 and b is positive even number, then \(a^b=1\) and \(a^{b-1}=-1\)
1-(-1)=2 i.e. RHS, therefore it works for all even positive values of B, therefore insufficient.

Please give +1 Kudos, if you liked the explanation

If a and b are different nonzero integers, what is the value of b?

(1) \(a^b = ab\)
b = 1
SUFFICIENT

(2) \(a^b - a^{b - 1} = 2\)
\(a^{b-1}(a-1) = 2\)
\(a^{b-1} = \frac{2}{a-1}\)
Since b depends on a & a is unknown
NOT SUFFICIENT

IMO A