Exponents - ACT Math

7/8 is being raised to the 5th power and to the 1/2 power at the same time. We multiply these to find n.

To solve, we use logarithms. We log both sides and get: log3 = log2x

which can be rewritten as log3 = xlog2

Then we solve for x: x = log 3/log 2 = 1.5850 . . .

You can change the form to

3_x_ = 27

x = 3

In order to set the exponents equal to each other and solve for t, there must be the same number raised to those exponents.

64 = (√2)n?

(√2)2 = 2 and 26 = 64, so ((√2)2)6= (√2)2*6 = (√2)12.

Thus, we now have (√2)12(t+1) = (√2)10t + 4.

12(t+1) = 10t + 4

12t + 12 = 10t + 4

2t + 12 = 4

2t = –8

t = –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 first step is to distribute the squared on the second term. (2a3)2 becomes 4a6 by multiplying the exponents (power raised to a power exponent rule) and squaring the 2. Then, combining like terms (i.e. combining coefficients, a's and b's) we obtain 12a8b5.

This requires us to remember the rules for multiplying and adding exponent variables.

(3y2)2 can be re-written as (3)2 x (y2)2 which yields 9y4

(4y)3 can be re-written as (4)3 x (y)3 which yields 64y3

adding the two yields

9y4 +64y3

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

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.

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

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

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 .

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).