GreenlightTestPrep wrote:
If \(b\) is a non-zero integer, how many different values of \(b\) satisfy the equation \(b^{(2b^2)} = b^{8}\)
A) One
B) Two
C) Three
D) Four
E) Five
I created this question to remind students that, when it comes to equations with variables in the exponents, there are three important provisos we must consider before we can conclude that two exponents are equal.
That is, if \(x^a = x^b\), then we can conclude that \(a = b\) AS LONG AS \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\).
For example, if we know that \(0^x = 0^3\), we can't then conclude that \(x = 3\), since there are infinitely many values of \(x\) that satisfy the equation. Let's begin by assuming \(b\) does not equal any of the forbidden numbers (i.e., \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\))
In that case, we can conclude that \(2b^2 = 8\)
Divide both sides of the equation by \(2\) to get: \(b^2 = 4\)
So, \(b = 2\) or \(b = -2\)
Now we need to check whether each of the forbidden numbers is a possible solution to the given equation.
We already said that
b is a non-zero integer, so we need not bother testing \(b = 0\).
What about \(b = 1\)?
Plug it in to get: \(1^{(2(1^2))} = 1^{8}\)
Simplify: \(1^{2} = 1^{8}\)
Works!!
So, \(b = 1\) is a solution
What about \(b = -1\)?
Plug it in to get: \((-1)^{(2(-1)^2)} = (-1)^{8}\)
Simplify: \((-1)^{2} = 1^{8}\)
Works!!
So, \(b = -1\) is a solution
So we have a total of four solutions: \(b = 2\), \(b = -2\), \(b = 1\) and \(b = -1\)
Answer: D
If you want to practice this question type, here’s a similar question:
https://gre.myprepclub.com/forum/what-is-s ... 20887.html