x^a=x^b; is a=b

[geetikab605]

Hi,

I am a little confused by a problem on the Manhattan CAT paper

If ab ≠ 0, does a = b?

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

I understand why (2) is NOT sufficient but why cant (1) by itself be sufficient

I can solve it like so:- x^a=x^b
=> x^a/x^b= 1
=> x^(a-b)=X^0
thus since the base is a common across; we can state that a-b=0
=> a=b

Just wondering if my reasoning is mathematically right?

[Sage Pearce-Higgins]

Unfortunately not. I'll take the practical route first and then explain the theory. I strongly encourage you to approach questions like this by testing cases, i.e. picking examples. Even if you're really advanced at algebra this strategy is often easier and more reliable. And if you don't like algebra it's good news, as testing cases offers a great way to avoid algebra.

Thinking about statement (1), we know that x^a= x^b. A simple example would be 3^2 = 3^2. This agrees with our data and gives the answer 'yes' to the question 'does a = b?'. However, what if x = 1? Then we could have 1^4 = 1^5 (that's true). Now we have the answer 'no' to the question 'does a = b?'. Takeaway: get acquainted with unusual results for exponents such as those involving 1 and 0.

Algebraically, your mistake was replacing 1 with x^0. For most numbers that's accurate, but you forgot to consider the case in which x is 1. In that case you could replace 1 with x^[anything!]. Also, be careful when dividing by unknown variables in equations - you need to make sure that the variable isn't 0. That's another algebra error you made in solving this equation.