Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.
_________________

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
_________________

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).
_________________

Why "Solving Exponential Equations with the Same Base" does not hold?
[#permalink]

Show Tags

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

______________________________

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?
[#permalink]

Show Tags

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

Give kudos if it helps

Posted from my mobile device

gmatclubot

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