When we have an expression raised to a power that, in turn, is raised (within parentheses) to another power, we can multiply the exponents and raise the base number to the result of this multiplication.

Question Types:

When we have an expression raised to a power that, in turn, is raised (within parentheses) to another power, we can multiply the exponents and raise the base number to the result of this multiplication.

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

This property is also concerning algebraic expressions.

Question 1

\( (3^5)^4= \)

Question 2

\( (6^2)^{13}= \)

Question 3

Solve the exercise:

\( (a^5)^7= \)

Question 4

\( (4^2)^3+(g^3)^4= \)

Question 5

\( ((y^6)^8)^9= \)

$(3^5)^4=$

To solve the exercise we use the power property:$(a^n)^m=a^{n\cdot m}$

We use the property with our exercise and solve:

$(3^5)^4=3^{5\times4}=3^{20}$

$3^{20}$

$(6^2)^{13}=$

We use the formula:

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

Therefore, we obtain:

$6^{2\times13}=6^{26}$

$6^{26}$

Solve the exercise:

$(a^5)^7=$

We use the formula:

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

and therefore we obtain:

$(a^5)^7=a^{5\times7}=a^{35}$

$a^{35}$

$(4^2)^3+(g^3)^4=$

We use the formula:

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

$(4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}$

$4^6+g^{12}$

$((y^6)^8)^9=$

We use the power rule of distributing exponents.

$(a^m)^n=a^{m\cdot n}$We apply it in the problem:

$\big((y^6)^8\big)^9=(y^{6\cdot8})^9=y^{6\cdot8\cdot9}=y^{432}$When we use the aforementioned rule twice, the first time for the inner parentheses in the first stage and the second time for the remaining parentheses in the second stage, in the last stage we calculate the result of the multiplication in the power exponent.

__Therefore, the correct answer is option b.__

$y^{432}$

Question 1

\( (a^4)^6= \)

Question 2

\( (2^2)^3+(3^3)^4+(9^2)^6= \)

Question 3

\( ((b^3)^6)^2= \)

Question 4

Solve the exercise:

\( (x^2\times3)^2= \)

Question 5

\( (4^x)^y= \)

$(a^4)^6=$

We use the formula

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

Therefore, we obtain:

$a^{4\times6}=a^{24}$

$a^{24}$

$(2^2)^3+(3^3)^4+(9^2)^6=$

We use the formula:

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

$(2^2)^3+(3^3)^4+(9^2)^6=2^{2\times3}+3^{3\times4}+9^{2\times6}=2^6+3^{12}+9^{12}$

$2^6+3^{12}+9^{12}$

$((b^3)^6)^2=$

We use the formula

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

Therefore, we obtain:

$((b^3)^6)^2=(b^{3\times6})^2=(b^{18})^2=b^{18\times2}=b^{36}$

$b^{36}$

Solve the exercise:

$(x^2\times3)^2=$

We have an exponent raised to another exponent with a multiplication between parentheses:

$(z\cdot t)^n=z^n\cdot t^n$This says that in a case where a power is applied to a multiplication between parentheses,the power is applied to each term of the multiplication when the parentheses are opened,

We apply it in the problem:

$(3x^2)^2=3^2(x^2)^2$With the second term of the multiplication we proceed carefully, since it is already in a power (that's why we use parentheses). The term will be raised using the power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$and we apply it in the problem:

$3^2(x^2)^2=9x^{2\cdot2}=9x^4$In the first step we raise the number to the power, and in the second step we multiply the exponent.

__Therefore, the correct answer is option a.__

$9x^4$

$(4^x)^y=$

Using the law of powers for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$We apply it in the problem:

$(4^x)^y=4^{xy}$__Therefore, the correct answer is option a.__

$4^{xy}$

Question 1

\( ((3^9)^{4x)^{5y}}= \)

Question 2

\( ((14^{3x})^{2y})^{5a}= \)

Question 3

\( (y^3\times x^2)^4= \)

Question 4

\( x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=\text{?} \)

Question 5

\( [(\frac{1}{7})^{-1}]^4= \)

$((3^9)^{4x)^{5y}}=$

We use the power rule for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$We apply this rule to the given problem:

$((3^9)^{4x})^{5y}= (3^9)^{4x\cdot 5y} =3^{9\cdot4x\cdot 5y}=3^{180xy}$In the first step we applied the previously mentioned power rule and removed the outer parentheses. In the next step we applied the power rule once again and removed the remaining parentheses. In the final step we simplified the resulting expression.

__Therefore, the correct answer is option b.__

$3^{180xy}$

$((14^{3x})^{2y})^{5a}=$

Using the power rule for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$We apply the rule to the given problem:

$((14^{3x})^{2y})^{5a}=(14^{3x})^{2y\cdot5a}=14^{3x\cdot2y\cdot5a}=14^{30xya}$In the first step we applied the aforementioned power rule and removed the outer parentheses. In the next step we again applied the power rule and removed the remaining parentheses.

In the final step we simplified the resulting expression,

__Therefore, through the rule of substitution (which is applied to the exponent of the power in the obtained expression) it can be concluded that the correct answer is answer D.__

$14^{30axy}$

$(y^3\times x^2)^4=$

We will solve the problem in two steps, in the first step we will use the power of a product rule:

$(z\cdot t)^n=z^n\cdot t^n$The rule states that the power affecting a product within parentheses applies to each of the elements of the product when the parentheses are opened,

We begin by applying the law to the given problem:

$(y^3\cdot x^2)^4=(y^3)^4\cdot(x^2)^4$When we open the parentheses, we apply the power to each of the terms of the product separately, but since each of these terms is already raised to a power, we must be careful to use parentheses.

We then use the power of a power rule.

$(b^m)^n=b^{m\cdot n}$We apply the rule to the given problem and we should obtain the following result:

$(y^3)^4\cdot(x^2)^4=y^{3\cdot4}\cdot x^{2\cdot4}=y^{12}\cdot x^8$When in the second step we perform the multiplication operation on the power exponents of the obtained terms.

__Therefore, the correct answer is option d.__

$y^{12}x^8$

$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=\text{?}$

**First** we will rearrange the expression and use the fact that multiplying a fraction means multiplying the numerator of the fraction, and the distributive property of multiplication:

$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=x^3\cdot x^4\cdot 2\cdot \frac{1}{x^3}\cdot x^{-8}=2\cdot x^3\cdot x^4\cdot \frac{1}{x^3}\cdot x^{-8}$Next, we'll use the law of exponents for negative exponents:

$a^{-n}=\frac{1}{a^n}$We'll apply the law of exponents to the expression in the problem:

$2\cdot x^3\cdot x^4\cdot\frac{1}{x^3}\cdot x^{-8} =2\cdot x^3\cdot x^4\cdot x^{-3}\cdot x^{-8}$When we applied the above law of exponents for the fraction in the multiplication,

**From now on, we will no longer use the multiplication sign and will switch to the conventional notation where juxtaposition of terms means multiplication between them,**

Now we'll recall the law of exponents for multiplying terms with the same base:

$a^m\cdot a^n=a^{m+n}$And we'll apply this law of exponents to the expression we got in the last step:

$2x^3x^4x^{-3}x^{-8} =2x^{3+4+(-3)+(-8)} =2x^{3+4-3-8}=2x^{-4}$When in the first stage we applied the above law of exponents and in the following stages we simplified the expression in the exponent,

Let's summarize the solution steps so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}= 2 x^3x^4 x^{-3} x^{-8} =2x^{-4}$

Now let's note **that there is no such answer in the given options**, a further check of what we've done so far will also reveal **that there is no calculation error**,

Therefore, we can conclude that **additional mathematical manipulation** is required to determine which is the correct answer among the suggested answers,

**Let's note** that in answers A and B there are **similar** expressions to the one we got in the last stage, however - we can directly rule out the other two options since they are clearly different from the expression we got,

Furthermore, we'll note that in the expression we got, x is in a negative exponent and is in the numerator (** Note at the end of the solution on this topic**), whereas in answer B it is in a positive exponent and in the denominator (and both are in the numerator -

**If so - we are left with only one option - which is answer A',** however we want to verify (and need to verify!) that this is

Let's note that in the expression we got x is in a negative exponent and is in the numerator (** Note at the end of the solution on this topic**), whereas in answer B it is in a positive exponent and in the denominator ,

In addition, let's note that in answer B x is in the second power but inside parentheses that are also in the second power, whereas in the expression we got in the last stage of solving the problem x is in the fourth power** which might remind us of the law of exponents for power to a power**,

We'll check this, starting with the law of exponents for negative exponents mentioned at the beginning of the solution, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Next, we'll represent the term with the negative exponent that we got in the last stage of solving the problem, as a term in the denominator of the fraction with a positive exponent:

$2x^{-4}=2\cdot\frac{1}{x^4}$When we applied the above law of exponents,

Next, let's note that using the law of exponents for power to a power, **but in the opposite direction**:

$a^{m\cdot n}= (a^m)^n$We can conclude that:

$x^4=x^{2\cdot2}=(x^2)^2$Therefore, we'll return to the expression we got in the last stage and apply this understanding:

$2\cdot\frac{1}{x^4} =2\cdot\frac{1}{(x^2)^2}$** Let's summarize then** the problem-solving stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4} =2\cdot\frac{1}{(x^2)^2}$**Let's note that we still haven't got the exact expression suggested in answer A**, but we are already very close,

To reach the exact expression claimed in answer A, we'll recall another important law of exponents, and a useful mathematical fact:

**Let's recall** the law of exponents for exponents applying to terms in parentheses, **but in the opposite direction**:

$\frac{a^n}{c^n}=\big(\frac{a}{c}\big)^n$And let's also recall the fact that **raising the number 1 to any power will yield the result 1**:

$1^{x}=1$And therefore we can write the expression we got in the last stage in the following way:

$2\cdot\frac{1}{(x^2)^2}=2\cdot\frac{1^2}{(x^2)^2}$And then **since in the numerator and denominator of the fraction there are terms with the same exponent** we can apply the above law of exponents, and represent the fraction whose numerator and denominator are **terms with the same exponent** as a fraction whose numerator and denominator are **the bases of the terms** and it is raised **to the same exponent**:

$2\cdot\frac{1^2}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$** Let's summarize** then the solution stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4}=2\cdot\frac{1}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$__And therefore the correct answer is indeed answer A.__

__Note:__

When it's written "the number in the numerator" despite the fact that there is no fraction in the expression at all, it's because we can always refer to any number as a number in the numerator of a fraction if we remember that any number divided by 1 equals itself, that is, we can always write a number as a fraction by writing it like this:

$X=\frac{X}{1}$And therefore we can actually refer to $X$as a number in the numerator of a fraction.

$2(\frac{1}{x^2})^2$

$[(\frac{1}{7})^{-1}]^4=$

We use the power property of a negative exponent:

$a^{-n}=\frac{1}{a^n}$We will rewrite the fraction in parentheses as a negative power:

$\frac{1}{7}=7^{-1}$Let's return to the problem, where we had:

$\bigg( \big( \frac{1}{7}\big)^{-1}\bigg)^4=\big((7^{-1})^{-1} \big)^4$We continue and use the power property of an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$And we apply it in the problem:

$\big((7^{-1})^{-1} \big)^4 =(7^{-1\cdot-1})^4=(7^1)^4=7^{1\cdot4}=7^4$__Therefore, the correct answer is option c__

$7^4$