How to find the exponent of variables - ISEE Upper Level Mathematics Achievement

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Question

Simplify:

$\left (6x ^{-5 } \right )^{-3}$

Answer

Apply the power of a product property:

$\left (6x ^{-5 } \right )^{-3}$

$= 6 ^{-3}\cdot \left (x ^{-5 } \right )^{-3}$

$=\frac{1}{216}\cdot x ^{-5 \cdot (-3) }$

$=\frac{1}{216}\cdot x ^{15 }$

$=\frac{x ^{15 }}{216}$

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Question

What is the coefficient of $x^{5}$ in the expansion of $(x+1)^{12}$ .

Answer

By the Binomial Theorem, if $(x + 1)^{n}$ is expanded, the coefficient of $x^{r}$ is

$_{n}\textrm{C} _{r} = \frac{n!}{(n-r)!r!}$ .

Substitute : The coefficient of $x^{5}$ is:

$_{12}\textrm{C} _{5} = \frac{12!}{(12-5)!5!} = \frac{12!}{7!5!}$

$= \frac{12 \cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

$= \frac{12 \cdot 11\cdot 10\cdot 9\cdot 8}{ 5\cdot 4\cdot 3\cdot 2\cdot 1} = \frac{11\cdot 9\cdot 8}{ 1} = 792$

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Question

What is the coefficient of $x^{8}$ in the expansion of $(x + 3)^{10}$ ?

Answer

By the Binomial Theorem, the $x^{r}$ term of $(x + b)^{n}$ is

$_{n}\textrm{C} _{r} b^{n-r} x^{r}= \frac{n!}{(n-r)!r!} b^{n-r} x^{r}$ .

Substitute and this becomes

$\frac{10!}{(10-8)!8!} 3^{10-8} x^{8}$ .

The coefficient is

$\frac{10!}{(10-8)!8!} \cdot 3^{10-8} = \frac{10!}{2!8!}\cdot 3^{2} = \frac{9 \cdot 10}{2}\cdot 9 = 405$ .

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Question

What is the coefficient of $x^{4}$ in the expansion of $(2x + 3)^{6}$ ?

Answer

By the Binomial Theorem, the $x^{r}$ term of $(ax + b)^{n}$ is

$_{n}\textrm{C} _{r} b^{n-r}\left ( ax \right )^{r}= \frac{n!}{(n-r)!r!} a^{r} b^{n-r} x^{r}$ ,

making the coefficient of $x^{r}$

$\frac{n!}{(n-r)!r!} a^{r} b^{n-r}$ .

We can set in this expression:

$\frac{6!}{(6-4)!4!} \cdot 2^{4} \cdot 3^{6-4}$

$=\frac{6!}{2!4!}\cdot 2^{4}\cdot 3^{2}$

$=\frac{720}{2\cdot 24}\cdot 16\cdot 9$

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Question

Simplify the expression: $\left [\left ( x ^{3} \right )^{3} \right ]^{3 }$

Answer

Apply the power of a power property twice:

$\left [\left ( x ^{3} \right )^{3} \right ]^{3 } = ( x ^{3; \cdot ; 3 } )^{3 } = ( x ^{9 } )^{3 } = x ^{9\cdot 3} =x ^{27}$

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Question

Evaluate:

$\left [ (x^4)^4 \right ]^5$

Answer

We need to apply the power of power rule twice:

$\left [ (x^4)^4 \right ]^5=(x^{4\times 4})^5=(x^{16})^5=x^{16\times 5}=x^{80}$

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Question

$(x^{-4})^{a}=x^{8}$

Solve for .

Answer

Based on the power of a product rule we have:

$(x^{-4})^{a}=x^{8}\Rightarrow x^{-4\times a}=x^8$

The bases are the same, so we can write:

$-4a=8\Rightarrow a=\frac{8}{-4}=-2$

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Question

Simplify:

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

Answer

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

$(\frac{2x^5}{3y^2})^{-4}=(\frac{3y^2}{2x^5})^4$

Apply the exponent within the parentheses and simplify.

$\frac{(3y^2)^4}{(2x^5)^4}$

$\frac{3^4(y^2)^4}{2^4(x^5)^4}$

$\frac{81y^{(2\times 4)}}{16x^{(5\times 4)}}$

$\frac{81y^8}{16x^{20}}$

This fraction cannot be simplified further.

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Question

Simplify:

$(\frac{x^2y}{2z^3})^{-2}$

Answer

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

$(\frac{x^2y}{2z^3})^{-2}=(\frac{2z^3}{x^2y})^2$

Apply the exponent within the parentheses and simplify.

$\frac{(2z^3)^2}{(x^2y)^2}$

$\frac{2^2(z^3)^2}{(x^2)^2y^2}$

$\frac{4z^{(3\times 2)}}{x^{(2\times 2)}y^2}$

$\frac{4z^6}{x^4y^2}$

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Question

Simplify if and .

$\frac{x^3y+2x^2y+xy}{xy(x+1)}$

Answer

Begin by factoring the numerator and denominator. can be factored out of each term.

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

can be canceled, since it appears in both the numerator and denomintor.

$\frac{x^2+2x+1}{x+1}$

Next, factor the numerator.

$\frac{(x^2+2x+1)}{x+1}=\frac{(x+1)(x+1)}{x+1}$

Simplify.

$\frac{(x+1)(x+1)}{x+1}=x+1$

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Question

$\frac{1}{3}+ \sqrt{y}= \frac{2}{5}$

Evaluate .

Answer

$\frac{1}{3}+ \sqrt{y}= \frac{2}{5}$

$\frac{1}{3}+ \sqrt{y}- \frac{1}{3}= \frac{2}{5} - \frac{1}{3}$

$\sqrt{y} = \frac{2}{5} - \frac{1}{3}$

$\sqrt{y} = \frac{3 \cdot 2}{3 \cdot 5} - \frac{1 \cdot 5}{3 \cdot 5}$

$\sqrt{y} = \frac{6}{1 5} - \frac{ 5}{15}$

$\sqrt{y} = \frac{6- 5}{15}$

$\sqrt{y} = \frac{1}{15}$

$y = (\sqrt{y}) ^{2}= \left (\frac{1}{15} \right )^{2} = \frac{1^{2}}{15 ^{2}} = \frac{1}{225 }$

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Question

$\sqrt{y}-\frac{1}{3} = \frac{2}{5}$

Evaluate .

Answer

To solve for the variable isolate it on one side of the equation with all of constants on the other side.

$\sqrt{y}-\frac{1}{3} = \frac{2}{5}$

First add one third to both sides.

$\sqrt{y}-\frac{1}{3} + \frac{1}{3} = \frac{2}{5} + \frac{1}{3}$

$\sqrt{y} = \frac{2}{5} + \frac{1}{3}$

Calculate a common denominator to add the two fractions.

$\sqrt{y} = \frac{3 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 5}{3 \cdot 5}$

$\sqrt{y} = \frac{6}{1 5} + \frac{ 5}{15}$

$\sqrt{y} = \frac{6+ 5}{15}$

$\sqrt{y} = \frac{11}{15}$

Square both sides to solve for y.

$y = (\sqrt{y}) ^{2}= \left (\frac{11}{15} \right )^{2} = \frac{11^{2}}{15 ^{2}} = \frac{121}{225 }$

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Question

$t ^{3} = -8$

$u^{9} = -12$

Evaluate $(tu) ^{9}$ .

Answer

By the Power of a Product Principle,

$(tu) ^{9} = t ^{9} \cdot u ^{9}$

Also, by the Power of a Power Principle,

$(t^{3} ) ^{3} =t ^{3 \cdot 3} = t^{9}$

Combining these ideas, then substituting:

$(tu) ^{9} = (t^{3} ) ^{3} \cdot u ^{9}$

$= (-8) ^{3} \cdot (-12)$

$= (-8^{3} ) \cdot (-12)$

$= (-512 ) \cdot (-12)$

$= +(512 \cdot 12)$

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Question

$a^{3} = -10$

Evaluate $(2a) ^{6}$ .

Answer

By the Power of a Power Principle,

$(a^{3} ) ^{2} =a ^{3 \cdot 2} = a ^{6}$

So

$a ^{6} = (a^{3} ) ^{2} = (-10) ^{2} =+ (10 ^{2} )= 100$

Also, by the Power of a Product Principle,

$(2a) ^{6} = 2^{6} \cdot a ^{6}$

$2 ^{6} = 64$ , so, substituting,

$(2a) ^{6} = 64 ( 100 ) = 6,400$ .

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Question

Simplify the following:

Answer

Simplify the following:

Let's recall the rules for distributing exponents.

We treat coefficients (like the 7) like regular numbers and raise them to the new exponent.

We deal with variables (like the t, h, and b) by multiplying their current exponent by the new exponent.

Doing so yields:

$7^3t^{63}h^{34}b^{3*9}$

Simplify to get:

$343t^{18}h^{12}b^{27}$

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