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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If ab ≠ 0, does a = b?

(1) x^a = x^b (2) x = x^2

In the original condition, there are 3 variables(x,a,b), which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation, and 1 more equation is needed, which is likely to make E the answer. When 1) & 2), if x=1, a=b=1 -> yes. If x=1, a=1, b=2, it is no and not sufficient. Thus, the answer is E.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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we have both possibilities a could equal b but not necessarily...we can see that x=1 because only 1=1^2....now if we know that x=1 then a and be can be any positive integers and condition x^a = x^b will still be satisfied...and they can also be equal to each other....of course x can be zero and then again both cases are possible same as for x=1, a and b can or not be equal
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x^a=x^b Why can we not get rid of the base x, as it is the same on both sides and say a=b?

Remember we can automatically equate the exponents of equal bases when that base does not equal 0, 1 or -1:

\(1^a = 1^b\), for any values of a and b (they are not necessarily equal); \((-1)^a = (-1)^b\), for any even values of a and b (they are not necessarily equal); \(0^a = 0^b\), for any non-zero a and b (they are not necessarily equal).
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Why "Solving Exponential Equations with the Same Base" does not hold?
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24 Apr 2018, 12:48

This question has been discussed before on the forums but I do not understand why the "Solving Exponential Equations with the Same Base" rule does not hold in this follow example:

If ab ≠ 0, does a=b?

(1) x^a = x^b (2) x = x^2

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RULE: Solving Exponential Equations with the Same Base: If B^M = B^N, then M = N.

FOR STATEMENT #1 Why doesn't this rule hold???? I understand if you set x=1, a=2, b=2 then a=b BUT if you set x=1, a=2, b=-2 then a≠b . I am trying to understand the logic why the above RULE doesn't apply in this particular case??

FYI: The solution to this problem according to the answer key is E based on the "plug-in numbers" technique described above.

This question has been discussed before on the forums but I do not understand why the "Solving Exponential Equations with the Same Base" rule does not hold in this follow example:

If ab ≠ 0, does a=b?

(1) x^a = x^b (2) x = x^2

______________________________

RULE: Solving Exponential Equations with the Same Base: If B^M = B^N, then M = N.

FOR STATEMENT #1 Why doesn't this rule hold???? I understand if you set x=1, a=2, b=2 then a=b BUT if you set x=1, a=2, b=-2 then a≠b . I am trying to understand the logic why the above RULE doesn't apply in this particular case??

FYI: The solution to this problem according to the answer key is E based on the "plug-in numbers" technique described above.

Re: Why "Solving Exponential Equations with the Same Base" does not hold?
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24 Apr 2018, 13:07

Top Contributor

coolo wrote:

RULE: Solving Exponential Equations with the Same Base: If B^M = B^N, then M = N.

Your rule (above) requires some provisos.

For example, if 0^M = 0^N, we can't necessarily conclude that M = N Likewise, if 1^M = 1^N, we can't necessarily conclude that M = N Likewise, if 1^M = (-1)^N, we can't necessarily conclude that M = N. For example, M = 1 and N = 2 satisfies the given equation, but M ≠ N

1) x^a=x^b -> x^a/x^b=1 -> x^(a-b)=1^1=1^0. Now, a-b=1 -> a ≠ b (No) and a-b=0 -> a = b (Yes). Not sufficient. 2) x = x^2 - Not sufficient as we don't know about a and b.

1)+2) putting x=x^2 in 1) x^2a=x^2b -> x^2a/x^2b -> x^(2a-2b)=1^1=1^0, so 2a-2b=1 -> a ≠ b (No) and 2a-2b=0 -> a=b (Yes)

I) x^a=x^b If x=zero Then equation holds but we can't say whether a=b or not Both cases hold So insufficient

II) x^2-x=0 x(x-1)=0 x=0 or x=1 Insufficient as no info about a And b

Combining both X can be zero or 1 in either case Equation holds and a and b can have any value without changing result of equation so not sufficient E is answer

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gmatclubot

Re: If ab ≠ 0, does a = b? &nbs
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12 May 2018, 05:58