How to multiply exponents - ACT Math

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Question

If〖7/8〗n= √(〖7/8〗5),then what is the value of n?

Answer

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.

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Question

If 64t+1 = (√2)10t + 4, what is the value of t?

Answer

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

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Question

Simplify: (x3 * 2x4 * 5y + 4y2 + 3y2)/y

Answer

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

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Question

Which of the following is equivalent to:

Answer

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.

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Question

Simplify the following expression:

(3y2)2 + (4y)3

Answer

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

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Question

(b * b4 * b7)1/2/(b3 * bx) = b5

If b is not negative then x = ?

Answer

Simplifying the equation gives b6/(b3+x) = b5.

In order to satisfy this case, x must be equal to –2.

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Question

Simplify ((x²)-2)-3

Answer

We are given an expression with a power to a power to a power. Using rules of exponents, we take the exponents and multiply each of them together.

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Question

Solve: 5x2 – 3y1 where x = 4, y = 5.

Answer

Substitute the values for x and y within the equation: 5(4)2 - 3(5)1. Proceed according to proper order of operations: 5(16) – 3(5). Therefore: 80-15= 65.

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Question

Which of the following is a value of x that satisfies

$2^{x+2}\cdot 2^{2x-3}=32$

Answer

This question incorporates properties of exponents. The best way to solve this problem is to establish the same base for all of the terms.

$2^{x+2}\cdot 2^{2x-3}=2^{5}$

Now use the properties of exponents to simplify the left side of the equation. When exponential terms with the same base are multiplied, the exponents are added.

$2^{x+2}\cdot 2^{2x-3}=2^{(x+2)+(2x-3)}=2^{3x-1}$

Now, with the left side simplified, set that equal to $2^{5}$ .

$2^{3x-1}=2^{5}$

Since each side has one term with the same base, simply set the exponents equal to each other and solve for x.

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Question

Find the value of x where:

$4^{x}\times2^{x+2}=1$

Answer

$ewline 4^{x}\times2^{x+2}=1 ewline 2^{2x}\times2^{x+2}=1 ewline 2^{2x}\times2^{x+2}=2^{0} ewline 2^{3x+2}=2^{0} ewline 3x+2=0 ewline 3x=-2 ewline x=\frac{-2}{3}$

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Question

Simplify the following expression:

$\frac{4(x^4)^5}{(2^{-1})^{-3}(x^{-10})^{-1}}$

Answer

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

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Question

Simplify the following expression:

4x2 * 5x3

Answer

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

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Question

Evaluate

(_x_3)2

Answer

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

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Question

Evaluate:

$y=\frac{3^{13}-9^{5}}{\left ( \frac{1}{27} \right )^{-3}}$

Answer

Can be simplified to:

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Question

Simplify: $\frac{x^{-2}y^{5}z^{-4}}{x^{4}y^{3}}$

Answer

To tackle this problem we must understand the concept of exponents in fractions and how to cancel and move them.

To move any variable or number from the numerator to the denominator or vice versa, you must negate the exponent. i.e. $x^{2}$ in the numerator would become $x^{-2}$ in the denominator. These two expression are equivalent. You should strive to make all exponents positive initially before applying the next rule to simplify.

Cancelling variables with a similar base is an easy way to simplify. Add or subtract the exponents depending on their relationship in a fraction.

Ex. $\frac{x}{x}=\frac{1}{1}$ or 1.

Ex. $\frac{x^{2}}{x}=x$ . --> this can be more easily understood if you break down the $x^{2}$ . $\frac{(x\cdot x)}{x}$ which then can be moved around to form, $\frac{x}{x}\cdot x$ . After the $\frac{x}{x}$ cancels to form 1, we have $1\cdot x$ or just . This can be applied for all numerical or abstract values of exponents for a given variable, such as , or .

Knowing these rules, we can tackle the problem.

To begin we will pick a variable to start with, thereby breaking down the problem into three smaller chunks. First we will start with the variable . $\frac{x^{-2}}{x^{4}}$ . Because the numerator has a negative exponent, we will move it down to the denominator: $\frac{1}{x^{2}\cdot x^{4}}$ . This simplifies to $\frac{1}{x^{6}}$ as multiplying any common variables with exponents is found by addition of the exponents atop the original variable. The variable part of this problem is $\frac{1}{x^{6}}$ .

We move to the section of the problem: $\frac{y^{5}}{y^{3}}$ . This is similar to our $\frac{x^{2}}{x}$ above, instead with larger numerical exponents. $\frac{y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y}=\frac{y\cdot y\cdot y}{y\cdot y\cdot y}\cdot (y\cdot y)$ . The $\frac{y\cdot y\cdot y}{y\cdot y\cdot y}$ section cancels, leaving us with $y\cdot y$ or $y^{2}$ .

Now to the section. We simply have $z^{-4}$ on top. Applying the first rule above, we just move it to the denominator with the switching of the sign. Our result is $\frac{1}{z^{4}}$ .

Combining all the sections together we have $\frac{1}{x^{6}}\cdot y^{2}\cdot \frac{1}{z^{4}}$ .

More beautifully written it looks like $\frac{y^{2}}{x^{6}z^{4}}$ .

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Question

Simplify:

$(6x^{2})^{3}\cdot x^{-7}\cdot 2x^{4}$

Answer

Remember, we add exponents when their bases are multiplied, and multiply exponents when one is raised to the power of another. Negative exponents flip to the denominator (presuming they originally appear in the numerator).

$(6x^{2})^{3}\cdot x^{-7}\cdot 2x^{4}$

$=\frac{216x^{6}\cdot 2x^{4}}{x^{7}}$

$=\frac{432x^{10}}{x^{7}}$

$=432x^{3}$

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Question

Which of the following expressions is equivalent to $(-3x^{6}x^{2}y^{3})^{3}$ ?

Answer

$(-3x^{6}x^{2}y^{3})^{3}$ is simplified by multiplying the exponent outside of the expression , with each number and variable inside the expression: $(-3)^{3}=-27$

$(x^{6})^{3}=x^{18}$

$(x^{2})^{3}=x^{6}$

$(y^{3})^{3}=y^{9}$

This gives you $-27x^{18}x^{6}y^{9}$

When variables with exponents are multiplied, you add their respective exponents together, so $x^{18}x^{6}=x^{(18+6)}=x^{24}$

Altogether, the expression is simplified to $-27x^{24}y^{9}$

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Question

Compare $5^{27}(850)$ and $5^{29}(34)$ .

Answer

To compare these expressions more easily, we'll change the first expression to have $5^{29}$ in front. We'll do this by factoring out 25 (that is, $\small 5^{2}$ ) from 850, then using the fact that $\small 5^{27}5^{2}=5^{29}$ .

$5^{27}(850)=5^{27}(2534)=5^{27}(5^234)$

When we combine like terms, we can see that $5^{27}(5^234)=5^{29}(34)$ . The two terms are therefore both equal to the same value.

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Question

Which of the following is equal to $(5^4*5^5)^{20}$ ?

Answer

$x^{a} \times x^{b}$ is always equal to $x^{a + b}$ ; therefore, 5 raised to 4 times 5 raised to 5 must equal 5 raised to 9.

$(5^45^5)^{20}=(5^{(4+5)})^{20}=(5^9)^{20}$

$(x^{a})^{b}$ is always equal to $x^{(ab)}$ . Therefore, 5 raised to 9, raised to 20 must equal 5 raised to 180.

$(5^9)^{20}=5^{(9*20)}=5^{180}$

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Question

Which of the following is equal to ?

Answer

First, multiply inside the parentheses: .

Then raise to the 7th power: $(30^7)^7=30^{49}$ .

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