# Question: x to the 7th Power

## Comment on x to the 7th Power

### i have a problem with the

i have a problem with the part where u say square root of say 49 is +7 and not -7. In data sufficiency question 14 in this module, your say sq root of x can be +ve or -ve. I am perplexed. plz help me out ### This comes down to notation.

This comes down to notation.
If we're told that x² = 49, then x = 7 or x = -7

However, if we're told that x = √49, then the square root NOTATION (√) specifically directs us to provide the POSITIVE value that, when squared, yields 49.

So, even though 7 and -7 both yield 49 when squared, the square root NOTATION (√) specifically directs us to provide the POSITIVE value. That is, √49 = 7

Does that help?

ya thanks

### I did it a little bit

I did it a little bit different. For statement 2, I have x^14= 4. Since it is an equation, whatever I do to one side, I do to the other. So I took square root to both sides yielding x^7=2. This did not yield two results, just one. Thus, I concluded statement 2 was sufficient. What was my mistake? ### I'll show you by way of

I'll show you by way of analogy.

Take the equation k² = 9. We can see here that EITHER k = 3 OR k = -3. In other words, either k = √9 or either k = -√9

More general: (something)² = 4. We can see here that EITHER something = 2 OR something = -2

For the video question, we have x^14 = 4
Rewrite as (x^7)² = 4
From the logic used in the previous case, we know that EITHER x^7 = 2 OE x^7 = -2

NOTE: Doing the same thing to both sides of an equation works when you stick with the basics: addition, subtraction, multiplication and division. Other operations (like square roots) can cause problems.

For example, if we have y² = 25, we know that either y = 5 or y = -5.

However, if we try to take the positive square root of each side, we get x = 5, which misses one of the solutions.

### Does this problem actually

Does this problem actually require any calculations?

From statement 1 we know that x is positive because when raised to an odd power, it yields a positive result.

From statement 2 we cannot determine the sign of x because x is raised to an even power, therefore it could be positive or negative.

If it's this simple why would we even attempt to do any calculations? ### Yes, we could also apply some

Yes, we could also apply some number sense (as you have done) to this question. That said, since we're asked to find the value of x^7, I wanted students to be 100% convinced that we can/cannot determine the value of x^x.