Applying Combined Exponents Rules: Identify the greater value

In this problem, we are asked to determine whether it's an equality or an inequality, and if it's an inequality - what is its direction?

To do this, we will first use the law of exponents for negative exponents:

$x^{-n}=\frac{1}{x^n}$Before we start solving the problem, let's understand this law in a slightly different way:

Note that if we treat this law as an equation (in fact it is an equation in every sense), and multiply both sides of the equation by the common denominator which is:

$x^n$We get:

$x^{-n}=\frac{1}{x^n}\\ \frac{x^n}{1} =\frac{1}{x^n}\hspace{8pt} \text{/}\cdot x^n\\ x^n\cdot x^{-n}=1$In the first part we recall that any number can be represented as itself divided by 1. We apply this to the left side of the equation, then we multiply by the common denominator.

To know by how much we need to multiply each numerator (after reduction with the common denominator) we ask the question "By how much did we multiply the current denominator to get the common denominator?".

Let's see the result we got:

$x^n\cdot x^{-n}=1$Meaning that $x^n,\hspace{4pt}x^{-n}$are reciprocal numbers to each other, or in other words:

$x^n$is reciprocal to $x^{-n}$(and vice versa),

And in particular:

$x,\hspace{4pt}x^{-1}$are reciprocal to each other,

We can apply this understanding to the problem if we also remember the fact that the reciprocal of a fraction is the number we get by swapping the numerator and denominator, meaning that the fractions:

$\frac{z}{w},\hspace{4pt}\frac{w}{z}$are reciprocal fractions to each other - which can be understood logically, as their multiplication will clearly give the result 1.

And if we combine this with the previous understanding, we can easily conclude that:

$\big(\frac{z}{w}\big)^{-1}=\frac{w}{z}$Meaning that raising a fraction to the power of negative one will give us the reciprocal fraction, obtained by swapping the numerator and denominator.

Let's return to the problem and apply these understandings, in addition we'll recall the law of multiplying exponents, but in the opposite direction:

$(z^m)^n=z^{m\cdot n}$

We'll also apply this law to the problem, we'll first deal with the left term:

$\big(\frac{a}{b}\big)^{-7}\cdot b^8$We'll start with the first term in the expression:

$\big(\frac{a}{b}\big)^{-7}= \big (\frac{a}{b} \big )^{-1\cdot 7}= \big (\big (\frac{a}{b}\big )^{-1} \big )^{7}$In the first part we present the exponent expression as a multiplication between two numbers, in the second part we apply the law of multiplying exponents in its opposite direction.

Next, we'll apply the understanding that raising a fraction to the power of negative one will always give the reciprocal fraction, obtained by swapping the numerator with the denominator: we'll apply this to the first term in the expression we got in the last part:

$\big (\big (\frac{a}{b}\big )^{-1} \big )^{7} = \big (\frac{b}{a} \big )^{7}$Let's summarize. We got that:

$\big(\frac{a}{b}\big)^{-7}\cdot b^8 = \big (\big (\frac{a}{b}\big )^{-1} \big )^{7}\cdot b^8= \big (\frac{b}{a} \big )^{7} \cdot b^8$

Now let's return to the problem and examine what we have:

$\big (\frac{b}{a} \big )^{7} \cdot b^8 \text{ }_{—\text{ }}\big(\frac{b}{a}\big)^7 \cdot b^{-4}$We use the distributive property and rearrange the right-side expression.

Note that, on both sides, the first expression (i.e., the fraction with the exponent) is identical. However, the second term in the multiplication is different on both sides, and this is because it's given that:

b>1 (If it could also be equal to one, we could argue that maybe these terms could be equal, but it's given that it's greater than one and therefore these terms are certainly different).

Therefore we can conclude that this is not an equality but an inequality, and we need to determine its direction.

Next, let's note that since it's also given that:

a>0 We can conclude that:

\big (\frac{b}{a} \big )^{7} >0 And this is because both the numerator of the fraction and the denominator of the fraction are positive numbers,

And therefore the direction of the inequality is not dependent on this term, (if we didn't know the sign of this term for certain, we wouldn't be able to determine the direction of the inequality later on)

Meaning-

The term that will determine the direction of the inequality is the second term in the multiplication on both sides, meaning- we need to find the direction between the terms:

$b^8 \text{ }_{—\text{ }}b^{-4}$We keep the original sides these terms were on.

It will be enough to answer the given problem.

For this, we'll remember the rules of inequality for exponential expressions, which simply state that the direction of inequality between exponential expressions with equal bases will be determined both by the value of the bases and by the exponents in the following way:

For a base greater than one, the direction of inequality between the exponential expressions will maintain the direction of inequality between the exponents, meaning- for a base: $x$, such that:

$$ x>1 (The base is always defined to be a positive number)

And exponents $z,\hspace{4pt}w$such that: z>w It holds that:

x^z>x^w

And for a base smaller than 1 and greater than 0, the direction of inequality between the exponential expressions will be opposite to the direction of inequality between the exponents, meaning- for a base: $x$, such that:

$$ 1 >x>0 (The base is always defined to be a positive number)

And exponents $z,\hspace{4pt}w$such that: z>w It holds that:

x^w >x^z

Let's return then to the problem:

We are required to determine the direction of inequality between the expressions:

$b^8 \text{ }_{—\text{ }}b^{-4}$From what's given in the problem b>1 Meaning greater than one, and therefore the direction of inequality between the expressions will be the same as the direction of inequality that between the exponents.

Therefore, we'll examine the exponents of the expressions in question here.

Since it's clear that:

8>-4 Then it holds that:

b^8 \text{ }>{\text{ }}b^{-4}

And therefore the correct answer is answer B.