Exponential Operations - ACT Math

If n=4, then 64(4/12)=64(1/3)=4. Then, 4=mŸ4(1+m)/(m+4). If 2 is substituted for m, then 4=2Ÿ4(1+2)/(2+4)=2Ÿ41/2=2Ÿ√4=2Ÿ2=4.

Let's do each of these separately:

x3 * 2x4 * 5y = 2 * 5 * x3 * x4 * y = 10 * x7 * y = 10x7y

4y2 + 3y2 = 7y2

Now, rewrite what we have so far:

(10x7y + 7y2)/y

There are several options for reducing this. Remember that when we divide, we can "distribute" the denominator through to each member. That means we can rewrite this as:

(10x7y)/y + (7y2)/y

Subtract the y exponents values in each term to get:

10x7 + 7y

The answer is (_x_10)/2. When an exponent is raised to another power, you multiply the exponents.

Since you are multiplying, you multiply the integers and add the exponents, giving you 20x5.

You can simplify it into (_x_3)(_x_3) = _x_6

Since the base is 5 for each term, we can say 2 + n =12. Solve the equation for n by subtracting 2 from both sides to get n = 10.

When you multiply exponents, you add the common bases:

y4 x7 + y5x6 + y18x4 + y3x26

Using the rules of exponents, 23 + 22 = 8 + 4 = 12

The only value of x where the two equations equal each other is 1. All you have to do is substitute the answer choices in for x.

Multiplying the two numbers yields 1080. Expressed in scientific notation 1080 is 1.08 x 103.

When common variables have exponents that are multiplied, their exponents are added. So _K_3 * _K_4 =K(3+4) = _K_7. And _M_6 * _M_2 = M(6+2) = _M_8. So the answer is _K_7/_M_8.

Add the coefficients of similar variables (y, y2, 9y3)

3y2 + 7y2 + 9y3 – y3 + y =

(3 + 7)y2 + (9 – 1)y3 + y =

10y2 + 8y3 + y

h–2n = 1/h2n

hn + h–2n = hn + 1/h2n

You multiply the integers, then add the exponents on the x's, giving you 24_x_5.

When multiplying exponents you smiply add the exponents.

For 2_x_² times 3_x_, 2 times 3 is 6, and 2 + 1 is 3, so 2_x_² times 3_x_ = 6_x_3

Rewrite the term on the left as a product. Remember that negative exponents shift their position in a fraction (denominator to numerator).

The term on the right can be rewritten, as 27 is equal to 3 to the third power.

Exponent rules dictate that multiplying terms allows us to add their exponents, while one term raised to another allows us to multiply exponents.

We now know that the exponents must be equal, and can solve for .

When multiplying bases that have exponents, simply add the exponents. Note that you can only add the exponents if the bases are the same. Thus:

First, reduce all values to a common base using properties of exponents.

Plugging back into the equation-

Using the formula

We can reduce our equation to

So,

The rule for adding exponents is . We can thus see that and are no more compatible for addition than and are.

You could combine the first two terms into , but note that PEMDAS prevents us from equating this to (the exponent must solve before the distribution).

Since the problem requires us to finish in a power of 2, it's easiest to begin by reducing all terms to powers of 2. Fortunately, we do not need to use logarithms to do so here.

Thus,