Applying Combined Exponents Rules: Using the laws of exponents

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 formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

$a^m\times a^n=a^{m+n}$

$3^2\times3^1=3^{2+1}=3^3$

We use the power property to multiply terms with identical bases:

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

$Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2$When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

In order to solve the given problem, we will follow these steps:

• Step 1: Simplify $2^0$. According to the zero exponent rule, $2^0 = 1$.

• Step 2: Simplify $3^{-4}$. Using the negative exponent rule, $3^{-4} = \frac{1}{3^4}$.

• Step 3: Simplify $9^2$. Recognize that $9 = 3^2$, thus $9^2 = (3^2)^2 = 3^{4}$.

• Step 4: Substitute the simplified terms back into the expression:

$\frac{2^0 \cdot 3^{-4}}{5^4 \cdot 9^2} = \frac{1 \cdot \frac{1}{3^4}}{5^4 \cdot 3^{4}}$

• Step 5: Simplify by combining like bases: since $3^{-4}$ in the numerator can be combined with $3^4$ in the denominator, you have:

$= \frac{1}{5^4 \cdot 3^{4+4}} = \frac{1}{5^4 \cdot 3^8}$

Therefore, the simplified expression is $\frac{1}{5^4 \cdot 3^8}$.

The problem requires simplification $\frac{9^2 \cdot 3^{-4}}{6^3}$ using exponent rules. Here’s a step-by-step guide to solving it:

• Step 1: Convert each term to powers of a common base.

  Notice that $9$ is $3^2$ and $6$ is $2 \times 3$. Hence:

  $9^2 = (3^2)^2 = 3^{4}$

  Therefore, the expression becomes $\frac{3^4 \cdot 3^{-4}}{(2 \cdot 3)^3}$.

• Step 2: Simplify the numerator.

  Using the exponent multiplication rule: $3^4 \cdot 3^{-4} = 3^{4 + (-4)} = 3^0 = 1$.

• Step 3: Expand the denominator.

  Calculate $(2 \cdot 3)^3$ by applying the distributive property: $(2 \cdot 3)^3 = 2^3 \cdot 3^3$.

• Step 4: Simplify the expression.

  After simplifying, the entire expression is $\frac{1}{2^3 \cdot 3^3}$.

  This simplifies further to $\frac{1}{6^3} = 6^{-3}$, because $2^3 \cdot 3^3 = (2 \cdot 3)^3 = 6^3$.

Therefore, the solution to the problem is $6^{-3}$.

Let's start by simplifying the second term in the complete multiplication, meaning - the fraction. We'll simplify it in two stages:

In the first stage we'll use the power law for multiplication between terms with identical bases:

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

and simplify the fraction's numerator:

$\frac{19^{35}\cdot19^{-32}}{19^3}=\frac{19^{35+(-32)}}{19^3}=\frac{19^{35-32}}{19^3}=\frac{19^3}{19^3}$

Next, we can either remember that dividing any number by itself gives 1, or use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ to get that:$\frac{19^3}{19^3}=19^{3-3}=19^0=1$

where in the last step we used the fact that raising any number to the power of 0 gives 1, meaning mathematically that:

$X^0=1$

Let's summarize this part, we got that:

$\frac{19^{35}\cdot19^{-32}}{19^3}=1$

Let's now return to the complete expression in the problem and substitute this result in place of the fraction:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}\cdot1=3^{-3}$

In the next stage we'll recall the power law for negative exponents:

$a^{-n}=\frac{1}{a^n}$

and apply this law to the result we got:

$3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

Summarizing all the steps above, we got that:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}=\frac{1}{27}$

Therefore the correct answer is answer A.

To solve this problem, we'll follow these steps:

• Step 1: Simplify $\sqrt[6]{b^{12}}$ using fractional exponent form.
• Step 2: Simplify the expression $\left(\frac{1}{b}\right)^2$ using negative exponents.
• Step 3: Combine and simplify the entire expression.

Now, let's work through each step:

Step 1: Simplify $\sqrt[6]{b^{12}}$.
The expression $\sqrt[6]{b^{12}}$ can be rewritten using fractional exponents as $b^{12/6} = b^2$.

Step 2: Simplify $\left(\frac{1}{b}\right)^2$.
The expression $\left(\frac{1}{b}\right)^2$ simplifies using negative exponents: $b^{-2}$.

Step 3: Combine the simplified expressions and the original variable $a$.
Combine all components as follows:
$b^2 \cdot b^{-2} \cdot a$.
Using the property $x^m \cdot x^n = x^{m+n}$, we have:
$b^{2 + (-2)} \cdot a = b^0 \cdot a$.
Since $b^0 = 1$ (by the zero exponent rule), the expression simplifies to:
$a$.

Therefore, the solution to the problem is $a$.

When raising any number to the power of 0 it results in the value 1, mathematically:

$X^0=1$

Apply this to both the numerator and denominator of the fraction in the problem:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1\cdot6^7}{36^4\cdot1}=\frac{6^7}{36^4}$

Note that -36 is a power of the number 6:

$36=6^2$

Apply this to the denominator to obtain expressions with identical bases in both the numerator and denominator:

$\frac{6^7}{36^4}=\frac{6^7}{(6^2)^4}$

Recall the power rule for power of a power in order to simplify the expression in the denominator:

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

Recall the power rule for division between terms with identical bases:

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

Apply these two rules to the expression that we obtained above:

$\frac{6^7}{(6^2)^4}=\frac{6^7}{6^{2\cdot4}}=\frac{6^7}{6^8}=6^{7-8}=6^{-1}$

In the first stage we applied the power of a power rule and proceeded to simplify the expression in the exponent of the denominator term. In the next stage we applied the second power rule - The division rule for terms with identical bases, and again simplified the expression in the resulting exponent.

Finally we'll use the power rule for negative exponents:

$a^{-n}=\frac{1}{a^n}$

We'll apply it to the expression that we obtained:

$6^{-1}=\frac{1}{6}$

Let's summarize the various steps of our solution:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1}{6}$

Therefore the correct answer is A.

To solve the problem, let's follow these steps:

• Step 1: Simplify the expression $45^{-80} \cdot \frac{1}{45^{-81}}$.
• Step 2: Simplify $49 \cdot 7^{-5}$.
• Step 3: Combine results to get the final expression.

Now, let's work through each step:
Step 1: Simplify $45^{-80} \cdot \frac{1}{45^{-81}}$. Using the exponent rule: $a^{-b} = \frac{1}{a^b}$ and $a^m \cdot a^n = a^{m+n}$, we have:

$45^{-80} \cdot \frac{1}{45^{-81}} = 45^{-80} \cdot 45^{81} = 45^{-80 + 81} = 45^1 = 45.$

Step 2: Simplify $49 \cdot 7^{-5}$. Note that $49 = 7^2$, so we can rewrite this as: $49 \cdot 7^{-5} = 7^2 \cdot 7^{-5} = 7^{2-5} = 7^{-3} = \frac{1}{7^3}.$

Step 3: Combine these results: $45 \cdot \frac{1}{7^3} = \frac{45}{7^3}.$

Therefore, the solution to the problem is $\frac{45}{7^3}$.

Let's use the law of exponents for negative exponents:

$a^{-n} = \frac{1}{a^n}$and apply this law to the problem:

$10^8+10^{-4}+(\frac{1}{10})^{-16}=10^8+\frac{1}{10^4}+(10^{-1})^{-16}$$$when we apply the above law of exponents to the second term in the sum, and the same law but in the opposite direction - we'll apply it to the fraction inside the parentheses of the third term in the sum,

Now let's recall the law of exponents for exponent of an exponent:

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

$10^8+\frac{1}{10^4}+(10^{-1})^{-16}=10^8+\frac{1}{10^4}+10^{(-1)\cdot(-16)}=10^8+\frac{1}{10^4}+10^{16}$when we apply this law to the third term from the left and then simplify the resulting expression,

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

$10^8+10^{-4}+(\frac{1}{10})^{-16}=10^8+\frac{1}{10^4}+(10^{-1})^{-16} =10^8+\frac{1}{10^4}+10^{16}$Therefore the correct answer is answer A.

In order to solve this problem, we'll follow these steps:

• Step 1: Simplify each component using exponent rules

• Step 2: Apply multiplication and division of powers

• Step 3: Simplify the combined expression

Now, let's work through each step:

Step 1: Simplify $(\frac{1}{2})^8$. Using the power of a fraction rule, we have:

$\left(\frac{1}{2}\right)^8 = \frac{1^8}{2^8} = \frac{1}{2^8} = 2^{-8}$

Step 2: Substitute back into the original expression:

$\frac{2^{-4} \cdot 2^{-8} \cdot 2^{10}}{2^3}$

Combine the terms in the numerator using the product of powers rule:

$2^{-4} \cdot 2^{-8} \cdot 2^{10} = 2^{-4 + (-8) + 10} = 2^{-2}$

Now the expression becomes:

$\frac{2^{-2}}{2^3}$

Apply the division of powers rule:

$\frac{2^{-2}}{2^3} = 2^{-2 - 3} = 2^{-5}$

Thus, the solution to the problem is $2^{-5}$.

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

We decompose the fraction inside of the parentheses:

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

We obtain:

$7^4\times8^3\times\frac{1^4}{7^4}$

We simplify the powers: $7^4$

We obtain:

$8^3\times1^4$

Remember that the number 1 in any power is equal to 1, thus we obtain:

$8^3\times1=8^3$

We'll use the law of exponents for negative exponents, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Let's apply this law to the problem:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1}$When we apply the above law to the second term from the left in the sum, and convert the fraction to a term with a negative exponent,

Next, we'll use the law of exponents for multiplying terms with identical bases:

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

$4^5-4^6\cdot4^{-1} =4^5-4^{6+(-1)}=4^5-4^{6-1}=4^5-4^{5}=0$When we apply the above law of exponents to the second term from the left in the expression we got in the last step, then we'll simplify the resulting expression,

Let's summarize the solution steps:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1} =4^5-4^{5}=0$

We got that the answer is 0,

Therefore the correct answer is answer A.

We'll use the power rule for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$and we'll simplify the second term on the left in the equation using it:
$5^3+5^{-3}\cdot5^3=5^3+5^{-3+3}=5^3+5^0=5^3+1$ where in the first stage we applied the mentioned rule to the second term on the left, then we simplified the expression with the exponent, and in the final stage we used the fact that any number raised to the power of 0 equals 1,

We didn't touch the first term of course since it was already simplified,

Therefore the correct answer is answer C.

First let's note that the number 9 is a power of the number 3:

$9=3^2$

Therefore we can immediately move to a unified base in the problem, in addition we'll recall the law of powers for negative exponents but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$

Let's apply this to the problem:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1}$

In the first term of the multiplication we replaced the number 9 with a power of 3, according to the relationship mentioned earlier, and simultaneously the third term in the multiplication we expressed as a term with a negative exponent according to the aforementioned law of exponents.

Now let's recall two additional laws of exponents:

a. The law of exponents for power of a power:

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

b. The law of exponents for multiplication between terms with equal bases:

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

Let's apply these two laws to the expression we got in the last stage:

$(3^2)^4\cdot3^{-8}\cdot3^{-1}=3^{2\cdot4}\cdot3^{-8}\cdot3^{-1}=3^8\cdot3^{-8}\cdot3^{-1}=3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

In the first stage we applied the law of exponents for power of a power mentioned in a', in the next stage we applied the law of exponents for multiplication of terms with identical bases mentioned in b', then we simplified the resulting expression.

Let's summarize the solution steps:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1} =3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

Therefore the correct answer is answer b'.

Begin by writing the problem and converting the decimal fraction in the problem to a simple fraction:

$\frac{10^4\cdot0.1^{-3}\cdot10^{-8}}{1000}=\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000}=\text{?}$

Next

a. We'll use the law of exponents for negative exponents:

$a^{-n}=\frac{1}{a^n}$

b. Note that the number 1000 is a power of the number 10:

$1000=10^3$

Apply the law of exponents from 'a' and the information from 'b' to the problem:

$\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000}=\frac{10^4\cdot(10^{-1})^{-3}\cdot10^{-8}}{10^3}$

We applied the law of exponents from 'a' to the term inside the parentheses of the middle term in the fraction's numerator. We applied the information from 'b' to the fraction's denominator,

Next, let's recall the law of exponents for power of a power:

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

And we'll apply this law to the same term we dealt with until now in the expression that we obtained in the last step:

$\frac{10^4\cdot(10^{-1})^{-3}\cdot10^{-8}}{10^3}=\frac{10^4\cdot10^{(-1)\cdot(-3)}\cdot10^{-8}}{10^3}=\frac{10^4\cdot10^3\cdot10^{-8}}{10^3}$

We applied the above law of exponents to the middle term in the numerator carefully, since the term in parentheses has a negative exponent. Hence we used parentheses and then proceeded to simplify the resulting expression,

Note that we can reduce the middle term in the fraction's numerator with the fraction's denominator. This is possible due to the fact that a multiplication operation exists between all terms in the fraction's numerator. Let's proceed to reduce:

$\frac{10^4\cdot10^3\cdot10^{-8}}{10^3}=10^4\cdot10^{-8}$

Let's summarize the various steps to our solution so far:

$\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000}=\frac{10^4\cdot(10^{-1})^{-3}\cdot10^{-8}}{10^3} = \frac{10^4\cdot10^3\cdot10^{-8}}{10^3} = 10^4\cdot10^{-8}$

Remember the law of exponents for multiplication of terms with identical bases:

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

Let's apply this law to the expression that we obtained in the last step:

$10^4\cdot10^{-8}=10^{4+(-8)}10^{4-8}=10^{-4}$

Now let's once again apply the law of exponents for negative exponents mentioned in 'a' above:

$10^{-4}=\frac{1}{10^4}=\frac{1}{10000}=0.0001$

When in the third step we calculated the numerical result of raising 10 to the power of 4 in the fraction's denominator. In the next step we converted the simple fraction to a decimal fraction,

Let's summarize the various steps of our solution so far:

$\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000} = \frac{10^4\cdot10^3\cdot10^{-8}}{10^3} = 10^{-4} =0.0001$

Therefore the correct answer is answer a.

Recall the law of exponents for multiplication between terms with identical bases:

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

We'll use this to deal with the fraction's denominator in the problem:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=\frac{(-3)^5\cdot8^4}{(-3)^{3+2+(-5)}}=\frac{(-3)^5\cdot8^4}{(-3)^0}$

In the first stage, we'll apply the above law to the denominator and then proceed to simplify the expression with the exponent in the denominator.

Remember that raising any number to the power of 0 gives the result 1, or mathematically:

$X^0=1$

Therefore the denominator that we obtain in the last stage is 1.

This means that:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=\frac{(-3)^5\cdot8^4}{(-3)^0}=\frac{(-3)^5\cdot8^4}{1}=(-3)^5\cdot8^4$

Recall the law of exponents for an exponent of a product inside of parentheses is as follows:

$(x\cdot y)^n=x^n\cdot y^n$

Apply this law to the first term in the product:

$(-3)^5\cdot8^4=(-1\cdot3)^5\cdot8^4 =(-1)^5\cdot3^5\cdot8^4=-1\cdot 3^5\cdot 8^4=-3^5\cdot8^4$

Note that the exponent applies separately to both the number 3 and its sign, which is the minus sign that is in fact multiplication by $-1$.

Let's summarize everything we did:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=(-3)^5\cdot8^4 = -3^5\cdot8^4$

Therefore the correct answer is answer C.