When multplying exponents, we need to make sure we have the same base.

Since we do, all we have to do is add the exponents.

We have two negative values so we treat as a regular addition problem. 

The answer is \(\displaystyle 10^{-19-17}=10^{-36}\). 

When multplying exponents, we need to make sure we have the same base.

Since we do, all we have to do is add the exponents.

Since there is a negative sign, we compare. \(\displaystyle 6\) is greater than \(\displaystyle 3\) and is positive so our answer must be positive. We treat the equation as a subtraction problem. 

The answer is \(\displaystyle 14^{-3+6}=14^{3}\). 

When multplying exponents, we need to make sure we have the same base.

Since we do, all we have to do is add the exponents. Since there is a negative sign, we compare. We can add all the negatie values which turns out to be \(\displaystyle -20\).\(\displaystyle 20\) is greater than \(\displaystyle 19\) and is negative so our answer must be negative.

We treat the equation as a subtraction problem. 

The answer is \(\displaystyle 2^{-20+19}=2^{-1}\). 

Although they have different bases, we can still simplify this expression. To get a base of \(\displaystyle 6\), we need a base of \(\displaystyle 2\) and \(\displaystyle 3\). Those are two numbers that give us a base of \(\displaystyle 6\). In order to change bases, both \(\displaystyle 2\) and \(\displaystyle 3\) bases must be raised to the same exponent. In this case they are. We can say \(\displaystyle 3^3*2^3\) is the same as \(\displaystyle 6^3\).

Now we have the same base, we can just add the exponents.

The answer is \(\displaystyle 6^{3+3}=6^6\). 

Although we have different bases, we can convert base \(\displaystyle 4\) into base \(\displaystyle 2\). \(\displaystyle 4\) is \(\displaystyle 2^2\). So for every \(\displaystyle 4\) we will always have \(\displaystyle 2^2\). We can set-up a proportion.

\(\displaystyle \frac{1}{2}=\frac{10}{x}\).

The top is power raised from base of \(\displaystyle 4\). The bottom is power raised from base of \(\displaystyle 2\). So now we cross multiply to get \(\displaystyle 20\). So now we have same base with different powers. We can add the exponents.

The answer is \(\displaystyle 2^{20+28}=2^{48}\). 

Although we have different bases, \(\displaystyle 27\) is basically \(\displaystyle 3*3*3\) or \(\displaystyle 3^3\).

We can set-up a proportion. 

\(\displaystyle \frac{1}{3}=\frac{x}{24}\) 

The top is the power raised from a base of \(\displaystyle 27\). The bottom is the power raised from a base of \(\displaystyle 3\). When we cross-multiply and divide both sides by \(\displaystyle 3\), we get \(\displaystyle 8\). So now we have a base of \(\displaystyle 27\) with different exponents.

We can add them up to get 

\(\displaystyle 27^{8+3}=27^{11}\).

We need to get them to bases of \(\displaystyle 2\). All of those bases derive from bases of \(\displaystyle 2\) but raised to different powers. \(\displaystyle 32\) is the same as \(\displaystyle 32^1\) as anything raised to the first power is the same as its base. \(\displaystyle 32=2^5\).

Next, \(\displaystyle 8=2^3\). Since it's raised to the fourth power, let's make a proportion. 

\(\displaystyle \frac{1}{3}=\frac{4}{x}\).
The top represents the power of base \(\displaystyle 8\). The bottom represents the power of base \(\displaystyle 2\). When we cross-multiply, we have \(\displaystyle 12\). So \(\displaystyle 8^4=2^{12}\).

Finally, \(\displaystyle 4\) is the same as \(\displaystyle 2^2\). Let's do another proportion.

\(\displaystyle \frac{1}{2}=\frac{2}{x}\). 

The top represents the power of base \(\displaystyle 4\). The bottom represents the power of base \(\displaystyle 2\).Wen we cross-mulltiply, we get \(\displaystyle 4\). So \(\displaystyle 4^2=2^4\).

With the same bases, we can add the exponents.

We have \(\displaystyle 2^{5+12+4}=2^{21}\). 

This will test your knowledge of a specific exponential property: \(\displaystyle (x^a)^b = x^{ab}\). Knowing that, 2 and 3 do come out to \(\displaystyle x^8\), while 1 comes out to only \(\displaystyle x^6\). 

When an exponent is raised to the power of another exponent, we multiply the exponents together.

For our x-value, the exponent will be \(\displaystyle \frac{1}{2}\cdot 2 =1\)

For our y-value, the exponent will be \(\displaystyle \frac{3}{2}\cdot 2=3\)

Therefore, 

\(\displaystyle (x^\frac{1}{2}y^\frac{3}{2})^2 = xy^3\)

\(\displaystyle t ^{- \frac{2}{5}} = h^{10}\)

\(\displaystyle \left (t ^{- \frac{2}{5}} \right )^{- \frac{5} {2}} =\left ( h^{10} \right )^{- \frac{5} {2}}\)

\(\displaystyle t ^{- \frac{2}{5} \cdot \left ( - \frac{5} {2} \right )} = h^{10 \cdot \left ( - \frac{5} {2} \right )}\)

\(\displaystyle t ^{1} = h^{-25 }\)

\(\displaystyle t = \frac{1}{h^{25}}\)