Lesson: Equations and Powers

Comment on Equations and Powers

P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
What is the value of P?
Sir when i am plugging in value as 0 it dosent satisfy the equation .So how is the answer 0?
Please help!

greenlight-admin's picture

x = 0 is, indeed, a solution to the equation.

Given: (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
Replace x with 0 to get: (0²)^(0² - 2(0) + 1) = 0^(3(0²) + 0 + 8)
Simplify: 0^1 = 0^8
Evaluate: 0 = 0

Since x = 0 is one of the solutions, the PRODUCT of all solutions will be 0.

Does that help?

Cheers,
Brent

Hello. For the first reinforcement activity, if I know the base cannot equal 0,1, or -1, why would I ever plug in those values? My answer was 4, because I thought we couldn’t use those values ever. Can you please explain why this was ok to do here? Thanks

Jason.
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the...

Hi Jason,

Be careful; we're not saying that the base cannot equal, 0, 1 or -1.

One of the rules says that, if b^x = b^y, then x = y. However, this rule does NOT hold true when b = 0, 1 or -1.
For example, if we know that 0^x = 0^y, we cannot conclude that x = y.

So, if we know that x^(2x² + 4x – 6) = x^(x² + 8x + 6), we cannot necessarily conclude that 2x² + 4x – 6 = x² + 8x + 6, since the original equation can also hold true when the base (x) is 0, 1 or -1.

So, when finding all of the solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x + 6), we must consider two different cases:

CASE A: the base (x) = 0, 1 or -1
CASE B: the base (x) does NOT equal 0, 1 or -1 (in which case 2x² + 4x – 6 = x² + 8x + 6)

As I show in my solution (at https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the...), x = 0 and x = 1 are two possible solutions when we explore CASE A.

Also, x = 6, and x = -2 are two more solutions when we explore CASE B.

Does that help?

Cheers,
Brent

Hi Brent,

For practice problem 1 & 2, I was not able to understand why do we check base value with either 0,1,-1. Bcz they are not valid for property to hold true.

For practice, problem 3 pls explain me this step

This means we can write: 2^(2y)/2^x = 20/5

How did we relate the two terms?

PLs help
greenlight-admin's picture

QUESTION: For practice problems 1 & 2, I was not able to understand why do we check base value with either 0, 1, -1. Bcz they are not valid for property to hold true.

We have a great rule that says: If b^x = b^y, then it must also be true that x = y
HOWEVER, this rule does NOT apply when b = 0, 1 or -1
So, if we want to apply the rule, we must always take the time to ensure that the base (b) does not equal 0, 1 or -1

For example, if 7^x = 7^y, then it is also true that x = y
However, if 0^x = 0^y, then we CANNOT conclude that x = y

So, if we have a VARIABLE for the base, we must be wary.
For example, if q^x = q^y, can we conclude that x = y?
No. In order to conclude that x = y, we must first ensure that q does not equal 0, 1 or -1

QUESTION: For practice, problem 3 pls explain me this step...
Here's question you're referring to (question #3): https://gre.myprepclub.com/forum/if-2-x-5-and-4-y-20-what-is-the-value-9...

Here's the principle I'm applying:
If w = x and y = z, then it's also true that w/y = x/z
For example, if x = 5 and y = 3, then x/y = 5/3

Likewise, in the question, we know that 2^(2y) = 20 and 2^x = 5
Applying above principle, we get: (2^2y)/(2^x) = 20/5

Does that help?

Cheers,
Brent

Hi, the first question from the "Reinforcement Activities" has:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6)?

In this type do we always need to consider the x = 0, -1 and 1 in addition to the roots of the equation

Thanks
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the...

Whenever we have equations with variables in the base, we must examine the possible special cases in which the base is 0, 1 or -1. Otherwise, we may make incorrect conclusions.

For example, if x^k = x^5, we cannot conclude that k = 5
For example, if x = 0, then we have 0^k = 0^5
As we can see, this equation will hold true for ALL values of k (not just k = 5)

Does that help?

Cheers,
Brent

https://gre.myprepclub.com/forum/if-2-n-3-3-n-2-1-36-what-is-the-value-of-n-7021.html
In this question i tried eliminating the denominator by multiplying both sides by 36 but ended up with a wrong answer. Can you solve out the attempt i made.
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-2-n-3-3-n-2-1-36-what-is-the-value-o...

I have a feeling that you're treating the left side as 2^(-n)/3 + 3^(-n)/2, when it is actually 2^(-n)/3 x 3^(-n)/2

So, let's first multiply the two fractions on the left side.
We get: [2^(-n) x 3^(-n)]/6 = 1/36

Multiply both sides by 36 to get: 6[2^(-n) x 3^(-n)] = 1

Since we have the same exponents we can combine them to get: 6[6^(-n)] = 1

Divide both sides by 6 to get: 6^(-n) = 1/6

Rewrite right side as follows: 6^(-n) = 6^(-1)

So, -n = -1

So, n = 1

Cheers,
Brent

I have to be more careful. I did treat it as addition instead of multiplication. Thanks brent

https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the-equation-3114.html

Hi Brent

In the question above you suggest 0^(-6)=0. Wouldn't it in fact be undefined?
greenlight-admin's picture

Solution link: https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the...

Good catch!!
You're right; 0^(-6) is undefined.
I've edited to my response accordingly.
Thanks for the heads up!

Cheers, Brent

https://gre.myprepclub.com/forum/if-9-x-1-2-2-2x-2-4-x-3-2x-3-then-what-is-the-value-21644.html
Hi, we are facing some issues to assume of factor out this sorts of questions. please provide a fully comprehensive and lucid explanation video at that kind of factor out
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-9-x-1-2-2-2x-2-4-x-3-2x-3-then-what-...
This is a very tricky (and time-consuming) question.

I believe the factoring you're referring to is when we factor 3^(2x-1) + 3^(2x-3) to get: 3^(2x-3)[3^2 +1]
The factoring strategy is the same here as it is for factoring x^7 + x^5 to get: x^5(x^2 + 1)

In both cases, the two terms have the same base (the base is 3 for the first expression, and the base is x for the second expression)

In both cases, we factor out other term with smaller exponent.
For the first expression, 2x-3 is the smaller exponent (since 2x-3 < 2x-1)
For the second expression, 5 is the smaller exponent (since 5 < 7)

Once we factor out 3^(2x-3) from the first expression, the first term in the brackets is 3^2 since 3^(2x-3) times 3^2 equals the first term in the original expression, 3^(2x -1)

Similarly, once we factor out x^5 from the second expression, the first term in the brackets is x^2 since x^5 times x^2 equals the first term in the original expression, x^7

Finally, when we expand 3^(2x-3)[3^2 +1], we get back to our original expression 3^(2x-1) + 3^(2x-3)
Similarly, when we expand x^5(x^2 + 1), we get back to our original expression x^5(x^2 + 1)

So, as you can see, the factoring for this question, albeit tricky, is the same as the greatest common factor factoring that is covered here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

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