# 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.

### Hi,

Hi,

For the 1st statement, I applied the rule n√X = X^1/n so for me X^21 = 8 can be rewritten as 3√X^21 = 3√8 then (X^21)^1/3 and now I have 2 powers that I can multiply to X^21/3 = X^7 = 2 ==> Sufficient

I followed the same logic for statement 2 but with square root as √X = X^1/2 and I found X^7 =2 as well.

I understand your approach but don't understand how my approach is wrong.

Thanks

### The main issue here is that

The main issue here is that odd roots preserve the sign (positive or negative) of the number we're taking the root of, whereas the even root notation "tells" us to output the POSITIVE root.

For example cuberoot(8) = 2. Notice that 2 and 8 are both positive (the sign is preserved).
Also notice that 2 is the ONLY number that, when raised to the power of 3, equals 8.

Similarly, cuberoot(-8) = -2. Notice that -2 and -8 are both negative
Also notice that -2 is the ONLY number that, when raised to the power of 3, equals -8.

Compare this to even roots.

Let's say we want to find the value of √25
In this case, there are TWO values (5 and -5) that, when raised to the power of 2, equal 25.
However, rather than allow √25 to have TWO values (5 and -5), mathematicians decided that the square root NOTATION will tell us to take only the positive root.

So, for example, if x² = 49, then x = √49 and x = -√49
The square root notation in √49 tells us to take the positive number that, when squared equals 49.
So, √49 = 7, which means -√49 = -7

Another example: k² = 9
We already know that this equation has two solutions (k = 3 and k = -3).
Now see what happens when we raise each side to the power of 1/2: (k²)^(1/2) = 9^(1/2)
This simplifies to be: k = √9
By imposing the square root NOTATION, we've suddenly restricted the TWO possible values of k to just the POSITIVE version.

The same thing applies with your solution.
When you took the cube root of each side (i.e., raise each side to the power of 1/3), we didn't lose any possible roots.
However, when you took the square root, we lost the negative root.

So, when raising a value to the power of 1/2 or 1/4 or 1/6 etc (i.e, taking an even root), we need to recognize that we'll likely be losing a possible root (the negative one).