Solve For Y: 8^y <= 16^(y+2)

less 16^{y+2}$

Hey guys, let's dive into a cool math problem today that's all about exponents and inequalities. We're going to tackle the question: If $8^y less 16^{y+2}$, what is the value of $y$? This might look a little intimidating with the exponents and the inequality sign, but trust me, it's totally manageable once we break it down. We'll be using some fundamental properties of exponents to make this a breeze. So, grab your thinking caps, and let's get this solved together! We're not just going to find the answer; we're going to understand why it's the answer. That's the best way to learn, right? We'll explore how to manipulate these exponential expressions, focusing on getting a common base, which is the golden ticket to solving these kinds of problems. Get ready to flex those math muscles because we're about to conquer this inequality!

Understanding the Basics: Powers and Bases

Alright, so the first thing we need to do when we see an inequality like $8^y less 16^{y+2}$ is to look at the bases. We have an 8 and a 16. These numbers aren't the same, which is a bit of a roadblock for directly comparing the exponents. The core strategy here is to express both bases using the same, smaller base. Think of it like translating two different languages into one common language so you can have a conversation. In the world of math, our common language for 8 and 16 is the number 2. Why 2, you ask? Because 8 is $2 imes 2 imes 2$, which is $2^3$, and 16 is $2 imes 2 imes 2 imes 2$, which is $2^4$. See? They both can be easily represented as powers of 2. This is super key, guys. If we can get both sides of the inequality to have the same base, we can then compare the exponents directly. This is a fundamental rule in algebra: if $a^m < a^n$ and $a > 1$, then $m < n$. Likewise, if $0 < a < 1$, then $m > n$. Since our base (2) is greater than 1, we'll use the first rule. So, let's get rewriting!

We'll substitute $8$ with $2^3$ and $16$ with $2^4$. Our inequality transforms from $8^y less 16^{y+2}$ into $(2^3)^y less (2^4)^{y+2}$. Now, we need to use another exponent rule: the power of a power rule. Remember that $(a^m)^n = a^{m imes n}$? This rule lets us multiply the exponents when we have a power raised to another power. Applying this to both sides of our inequality:

• For the left side: $(2^3)^y = 2^{3 imes y} = 2^{3y}$
• For the right side: $(2^4)^{y+2} = 2^{4 imes (y+2)} = 2^{4y + 8}$

So, our inequality now looks much simpler: $2^{3y} less 2^{4y + 8}$. We've successfully achieved our goal of having the same base (which is 2) on both sides. This is where the real problem-solving begins because now we can directly compare the exponents.

Manipulating the Exponents: Isolating 'y'

Now that we have the inequality in the form $2^{3y} less 2^{4y + 8}$, and knowing that the base (2) is greater than 1, we can drop the bases and focus solely on the exponents. This means the inequality $3y less 4y + 8$ is equivalent to the original one. Our mission, should we choose to accept it, is to isolate y on one side of this inequality. It's like a puzzle where you have to get all the 'y' pieces together.

To start, let's move all the y terms to one side. I usually prefer to move them to the side where they'll end up positive, if possible, to avoid negative signs causing confusion. Let's subtract $3y$ from both sides of the inequality:

$3y - 3y less 4y + 8 - 3y$

This simplifies to:

$0 less y + 8$

Now, we want to get y all by itself. To do that, we need to get rid of that '+ 8' on the right side. We can do this by subtracting 8 from both sides of the inequality:

$0 - 8 less y + 8 - 8$

This leaves us with:

$-8 less y$

Alternatively, we can write this as $y > -8$. This inequality tells us that for the original statement $8^y less 16^{y+2}$ to be true, the value of y must be greater than -8. So, any number larger than -8 will satisfy the condition. For example, if y=0, $8^0 = 1$ and $16^{0+2} = 16^2 = 256$. Clearly, $1 less 256$. If y=-7, $8^{-7}$ and $16^{-7+2} = 16^{-5}$. Since 8 and 16 are positive, their negative powers will be small fractions. We need to check if $8^{-7} less 16^{-5}$. Let's use our common base again: $(2^3)^{-7} = 2^{-21}$ and $(2^4)^{-5} = 2^{-20}$. Since $-21 < -20$, the inequality $2^{-21} less 2^{-20}$ is true. This confirms our solution. What if y = -8? Then we'd have $3(-8) = -24$ and $4(-8) + 8 = -32 + 8 = -24$. So, $2^{-24} less 2^{-24}$ is false. This is exactly why y must be strictly greater than -8.

Final Answer and Verification

So, after all that algebraic maneuvering, we've arrived at our solution: y > -8. This means any value of y that is greater than negative eight will make the original inequality $8^y less 16^{y+2}$ true. It's important to note that y cannot be equal to -8, because the inequality sign is strictly 'less than' ($less$), not 'less than or equal to' ($leqslant$).

Let's do a quick verification with a value greater than -8, say $y = -7$.

• Left side: $8^{-7} = (2^3)^{-7} = 2^{-21}$
• Right side: $16^{(-7)+2} = 16^{-5} = (2^4)^{-5} = 2^{-20}$

Is $2^{-21} less 2^{-20}$? Yes, it is, because -21 is indeed less than -20. So, $y = -7$ works.

Now let's try a value equal to -8, $y = -8$.

• Left side: $8^{-8} = (2^3)^{-8} = 2^{-24}$
• Right side: $16^{(-8)+2} = 16^{-6} = (2^4)^{-6} = 2^{-24}$

Is $2^{-24} less 2^{-24}$? No, this is false. $2^{-24}$ is equal to $2^{-24}$, not less than it. This confirms that $y = -8$ is not included in our solution set.

Finally, let's try a value less than -8, say $y = -9$.

• Left side: $8^{-9} = (2^3)^{-9} = 2^{-27}$
• Right side: $16^{(-9)+2} = 16^{-7} = (2^4)^{-7} = 2^{-28}$

Is $2^{-27} less 2^{-28}$? No, this is false. -27 is greater than -28. So, $y = -9$ does not satisfy the inequality.

This comprehensive verification solidifies our answer: y > -8. Great job working through this problem, guys! Remember, the key was finding a common base and then carefully manipulating the inequality. Keep practicing these exponent rules, and you'll be a pro in no time!