Power of a Quotient Rule for Exponents: Using multiple rules

First we rewrite the first expression on the left of the problem as a fraction:

$\frac{a^2}{a}+a^3\cdot a^5$Then we use two properties of exponentiation, to multiply and divide terms with identical bases:

1.

$b^m\cdot b^n=b^{m+n}$

2.

$\frac{b^m}{b^n}=b^{m-n}$

Returning to the problem and applying the two properties of exponentiation mentioned earlier:

$\frac{a^2}{a}+a^3\cdot a^5=a^{2-1}+a^{3+5}=a^1+a^8=a+a^8$

Later on, keep in mind that we need to factor the expression we obtained in the last step by extracting the common factor,

Therefore, we extract from outside the parentheses the greatest common divisor to the two terms which are:

$a$ We obtain the expression:

$a+a^8=a(1+a^7)$ when we use the property of exponentiation mentioned earlier in A.

$$$a^8=a^{1+7}=a^1\cdot a^7=a\cdot a^7$

Summarizing the solution to the problem and all the steps, we obtained the following:

$\frac{a^2}{a}+a^3\cdot a^5=a(1+a^{7)}$

Therefore, the correct answer is option b.

We use the three appropriate power properties to solve the problem:

1. Power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$ 2. Power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$ 3. Power law for the division of terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

We continue and apply the three previous laws to the problem:

$2^3\cdot2^4+(4^3)^2+\frac{2^5}{2^3}=2^{3+4}+4^{3\cdot2}+2^{5-3}=2^7+4^6+2^2$

In the first step we apply the power law mentioned in point 1 to the first expression on the left, the power law mentioned in point 2 to the second expression on the left, and the power law mentioned in point 3 to the third expression on the left, separately. In the second step, we simplify the expressions by exponents possession of the received terms,

Then,after using the substitution property for addition, we find that the correct answer is D.

In order to solve the problem we must use two power laws, as shown below:

A. Power property for terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$B. Power property for an exponent raised to another exponent:

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

We will apply these two power laws to the problem in two steps:

Let's start by applying the power law specified in A to the second term from the left in the given problem:

$\frac{2^7}{2^4}=2^{7-4}=2^3$In the first step we apply the power law specified in A and then proceed to simplify the resulting expression,

We then advance to the next step and apply the power law specified in B to the third term from the left in the given problem :

$(8^2)^5=8^{2\cdot5}=8^{10}$In the first stage we apply the power law specified in B and then proceed to simplify the resulting expression,

Let's summarize the two steps listed above to solve the general problem:

$(4\cdot7)^9+\frac{2^7}{2^4}+(8^2)^5= (4\cdot7)^9+2^3+8^{10}$In the final step, we calculate the result of multiplying the terms within the parentheses in the first term from the left:

$(4\cdot7)^9+2^3+8^{10}=28^9+2^3+8^{10}$Therefore, the correct answer is option c.

Let us begin by addressing the given problem as an equation:

$\big( \big(\frac{1}{5} \big)^2 \big)^?:5=125$Therefore, we shall replace the question mark with an x and proceed to solve it:

$\big( \big(\frac{1}{5} \big)^2 \big)^x:5=125$ Remember that dividing by a certain number is equivalent to multiplying by its inverse, so we will rewrite the given equation bearing this in mind:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$

Let's briefly discuss the solution technique:

Generally speaking the goal when solving exponential equations is to achieve a situation where there is a term on each of the two sides of the equation so that both sides have the same base. In such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and thus solve a simple equation for the unknown.

Mathematically, we will perform a mathematical manipulation (according to the laws of equation manipulation) on both sides of the equation. Or we will concentrate on the development of one of the sides of the equation with the help of power rules and algebra in order to reach the following situation:

$b^{m(x)}=b^{n(x)}$when $m(x),\hspace{4pt}n(x)$Algebraic expressions ( functions of the unknown $x$) that can also exclude the unknowns ($x$) that we are trying to find in the problem, which is the solution to the equation,

It is then stated that:

$m(x)=n(x)$and we solve the simple equation that we obtained.

We return to solving the equation in the given problem:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$In solving this equation, various power rules are used:

a. Power property with negative exponent:

$a^{-n}=\frac{1}{a^n}$b. Power property for a power of an exponent raised to another exponent:

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

Our initial goal is to simplify the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)

To do this, we will start by dealing with the fraction on the left side of the equation:

$\frac{1}{5}$That is, both the fraction inside the parenthesis and the fraction outside the parenthesis, this is done with the help of the power rule for a negative exponent specified in A above. We then represent this fraction as a term with a negative power and in the next step we apply the power rule for a power of an exponent raised to another exponent specified in B above. We then are able to remove the parentheses starting from the inner parenthesis to the outer ones. This is shown below step by step:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ (5^{(-1)\cdot 2} )^x\cdot 5^{-1}=125 \\ 5^{(-1)\cdot 2\cdot x} \cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\$When we carry out the development of the left side of the equation as described above, we initially apply the power rule for a negative exponent mentioned above in A.

In the following steps we apply the power rule for a power of an exponent raised to another exponent as mentioned above in B. We remove the parentheses: starting from the inner parenthesis to the outer. In the last step we simplify the expression in the power exponent on the left side of the equation,

c. Later we remember the power property for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Thus we apply this law to the left side of the equation that we obtained in the last step.

$5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x+(-1)}=125 \\ 5^{-2x-1}=125 \\$In the first step we apply the aforementioned power law to the product between members with identical bases mentioned above in C and in the following steps we simplify the expression in the power exponent on the left side,

Next, we seek to obtain the same base on both sides of the equation, the best way to achieve this is by decomposing each of the numbers in the problem into prime factors (using powers as well), you will notice that the number 125 is a power of the number 5, that is:

$125=5^3$This is the presentation (factorization) of the number 125 using its prime factor, which is the number 5.

So we return to the equation we obtained in the previous step and replace this number with its decomposition into prime factors:

$5^{-2x-1}=125 \\ 5^{-2x-1}=5^3 \\$We have reached our goal, we have obtained an equation in which both sides have terms with the same base, therefore we can state that the power exponents of the terms on both sides are equal, and in order solve the resulting equation for the unknown, we proceed as follows:

$5^{-2x-1}=5^3 \\ \\ \downarrow\\ -2x-1=3$We will continue to solve the resulting equation by isolating the unknown on the left side. We can achieve this in an usual way, by moving the sections and dividing the final equation by the unknown's coefficient:

$-2x-1=3 \\ -2x=3+1\\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \frac{\not{-2}x}{\not{-2}}=\frac{4}{-2}\\ x=-\frac{4}{2}\\ \bm{x=-2 }$In the first step we simplify the equation by moving the sides, remembering that when a term is moved its sign changes, then we complete the isolation by nullifying dividing both sides of the equation by its coefficient. In the last steps, we simplify the expression obtained by reducing the fractions,

We have thus solved the given equation. Below is a brief step by step summary of the solution:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x-1}=5^3 \\ \downarrow\\ -2x-1=3 \\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \bm{x=-2 }$Therefore, the correct answer is option a.