In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 4^{-8}\times 4^{6}=4^{(-8+6)}$

Solve for the exponents

$\displaystyle -8+6=-2$

$\displaystyle 4^{-2}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{4^{2}}=\frac{1}{16}$

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 2^{-8}\times 2^{3}=2^{(-8+3)}$

Solve for the exponents

$\displaystyle -8+3=-5$

$\displaystyle 2^{-5}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{2^{5}}=\frac{1}{32}$

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 3^{-6}\times 3^{2}=3^{(-6+2)}$

Solve for the exponents

$\displaystyle -6+2=-4$

$\displaystyle 3^{-4}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{3^{4}}=\frac{1}{81}$

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 3^{-21}\times 3^{16}=3^{(-21+16)}$

Solve for the exponents

$\displaystyle -21+16=-5$

$\displaystyle 3^{-5}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{3^{5}}=\frac{1}{243}$

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 5^{-19}\times 5^{17}=5^{(-19+17)}$

Solve for the exponents

$\displaystyle -19+17=-2$

$\displaystyle 5^{-2}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{5^{2}}=\frac{1}{25}$

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 9^{-12}\times 9^{9}=9^{(-12+9)}$

Solve for the exponents

$\displaystyle -12+9=-3$

$\displaystyle 9^{-3}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{9^{3}}=\frac{1}{729}$

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

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

Let's apply this rule to our problem

$\displaystyle 2^{-13}\times 2^{10}=2^{(-13+10)}$

Solve for the exponents

$\displaystyle -13+10=-3$

$\displaystyle 2^{-3}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$\displaystyle a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\displaystyle \frac{1}{2^{3}}=\frac{1}{8}$

Four times ten to the fifth power can be written numerically as the following:

$\displaystyle 4\times10^5$

When a number between one and ten is multiplied by a power of ten it is said to be written in scientific notation. This number is currently written in scientific notation.

Even though this appears to be a challenging math problem (i.e. because we have a power of ten), we can simply move our decimal place after the four, or $\displaystyle 4.0$, over $\displaystyle 5$ spaces to the right using zeros as place holders. 

$\displaystyle 4.0\times 10^5=400000.0$

Add commas and simplify.

$\displaystyle 400000.0\rightarrow400\textup,000$

Three times ten to the seventh power can be written numerically as the following:

$\displaystyle 3\times10^7$

When a number between one and ten is multiplied by a power of ten it is said to be written in scientific notation. This number is currently written in scientific notation.

Even though this appears to be a challenging math problem (i.e. because we have a power of ten), we can simply move our decimal place after the three, or $\displaystyle 3.0$, over $\displaystyle 7$ spaces to the right using zeros as place holders. 

$\displaystyle 3.0\times 10^7=30000000.0$

Add commas and simplify.

$\displaystyle 30000000.0\rightarrow30\textup,000\textup,000$

Two times ten to the eighth power can be written numerically as the following:

$\displaystyle 2\times10^8$

When a number between one and ten is multiplied by a power of ten it is said to be written in scientific notation. This number is currently written in scientific notation.

Even though this appears to be a challenging math problem (i.e. because we have a power of ten), we can simply move our decimal place after the two, or $\displaystyle 2.0$, over $\displaystyle 8$ spaces to the right using zeros as place holders. 

$\displaystyle 2.0\times 10^8=200000000.0$

Add commas and simplify.

$\displaystyle 200000000.0\rightarrow200\textup,000\textup,000$