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ISEE Upper Level Math : Equations

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Example Questions

Example Question #121 : Equations

Solve for :

$3^{4x-2}=1$

Possible Answers:

x=1

x=0.6

x=0

x=0.4

x=0.5

Correct answer:

x=0.5

Explanation:

We know that any number raised to the power of zero is equal to . Then we can substitute with 3^0 :

$3^0=1\Rightarrow3^{4x-2}=1=3^0$

Now we need to solve the following equation:

$3^{4x-2}=3^0$

Since the bases are the same, we can set the exponents equal to one another:

$4x-2=0\Rightarrow 4x=2\Rightarrow x=\frac{2}{4}=\frac{1}{2}=0.5$

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Example Question #122 : Equations

If is a real number and $y\neq 0$ , solve the following equation for :

$7^{2x-3}=y^0$

Possible Answers:

x=2

x=1.5

x=0.5

x=2.5

x=1

Correct answer:

x=1.5

Explanation:

We know that any number raised to the power of zero is equal to :

y^0=1

Then:

$7^{2x-3}=y^0=1$

Again we know that any number raised to the power of zero is equal to , so we can substitute with 7^0 :

$7^{2x-3}=1=7^0\Rightarrow 7^{2x-3}=7^0$

Since the bases are the same, we can set the exponents equal to one another:

$2x-3=0\Rightarrow 2x=3\Rightarrow x=\frac{3}{2}=1.5$

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Example Question #123 : Equations

Solve the following equation $(x\neq 2)$ :

$4^{\frac{x+2}{x-2}}=8$

Possible Answers:

x=6

x=4

x=8

x=7

x=10

Correct answer:

x=10

Explanation:

We know that 4=2^2 and 8=2^3 .

$4^{\frac{x+2}{x-2}}=8\Rightarrow \left [ 2^2 \right ]^{\frac{x+2}{x-2}}=2^3\Rightarrow 2^{2(\frac{x+2}{x-2})}=2^3$

Since the bases are the same, we can set the exponents equal to one another:

$2(\frac{x+2}{x-2})=3\Rightarrow (\frac{2x+4}{x-2})=3$

$\Rightarrow 2x+4=3(x-2)\Rightarrow 2x+4=3x-6$

$\Rightarrow 4+6=3x-2x\Rightarrow x=10$

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Example Question #121 : Equations

If $x\neq 3$ , solve the following equation for :

$5^{(\frac{2x-8}{x-3})}=1$

Possible Answers:

x=6

x=1

x=0

x=4

x=2

Correct answer:

x=4

Explanation:

We know that any number raised to the power of zero is equal to . Therefore we can substitute with 5^0 in the equation:

$5^{(\frac{2x-8}{x-3})}=5^0$

Since the bases are the same, we can set the exponents equal to one another:

${(\frac{2x-8}{x-3})}=0\Rightarrow 2x-8=0\Rightarrow 2x=8\Rightarrow x=\frac{8}{2}=4$

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Example Question #122 : Equations

If $x\neq 2$ , solve the following equation:

$3^{\frac{x^2-5}{x-2}}=27\times 3^x$

Possible Answers:

x=6

x=1

x=0

x=3

x=4

Correct answer:

x=1

Explanation:

We know that 27=3^3 :

$27\times 3^x=3^3\times 3^x\Rightarrow 27\times 3^x=3^{3+x}$

$3^{\frac{x^2-5}{x-2}}=27\times 3^x\Rightarrow 3^{\frac{x^2-5}{x-2}}=3^{3+x}$

Since the bases are the same, we can set the exponents equal to one another:

$\frac{x^2-5}{x-2}=3+x\Rightarrow x^2-5=(x-2)(3+x)$

$\Rightarrow x^2-5=3\times x+x\times x-2\times 3-2\times x$

$\Rightarrow x^2-5=3x+x^2-6-2x$

$\Rightarrow x^2-5=x^2+x-6$

$\Rightarrow x^2-x^2-5+6=x$

$\Rightarrow x=1$

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Example Question #122 : Equations

Define

$f(x) = x^{2}$

and

$g (x) = \left\{\begin{matrix} x+6\textrm{ if }x<0\\ x-4\textrm{ if }x \geq 0 \end{matrix}\right.$ .

Evaluate:

$\left ( f \circ g\right ) (0)$

Possible Answers:

Correct answer:

Explanation:

$\left ( f \circ g\right ) (0) = f(g(0))$

First, evaluate g(0) by using the definition of for nonnegative values of .

g(x) = x- 4

g(0) = 0- 4 = -4

Therefore, $\huge \left ( f \circ g\right ) (0) = f(g(0)) = f (-4)$

$f(x) = x^{2}$ , so

$f (-4) = (-4)^{2} = 16$

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Example Question #123 : Equations

Define $g (x) =8 - \frac{9}{10} x$

Evaluate g (3)

Possible Answers:

$6 \frac{3 }{10}$

$5 \frac{7 }{10}$

$6 \frac{7 }{10}$

$5 \frac{3 }{10}$

Correct answer:

$5 \frac{3 }{10}$

Explanation:

$g (x) =8 - \frac{9}{10} x$

$g (3) =8 - \frac{9}{10} \cdot 3$

$g (3) =8 - \frac{27}{10}$

$g (3) = \frac{80 }{10}- \frac{27}{10}$

$g (3) = \frac{53 }{10}$

$g (3) = 5 \frac{3 }{10}$

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Example Question #122 : Algebraic Concepts

Solve for :

7x - 17 > -5

Possible Answers:

$x> 3 \frac{1}{7}$

$x > -3 \frac{1}{7}$

$x > 1\frac{5}{7}$

$x >- 1\frac{5}{7}$

Correct answer:

$x > 1\frac{5}{7}$

Explanation:

7x - 17 >-5

7x - 17 +17> -5+17

7x >12

$7x\div 7 >12 \div 7$

$x > 1 \frac{5}{7}$

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Example Question #122 : Algebraic Concepts

Define $f (x) = x^{2} - 10$ and $g(x) = \sqrt{x+9}$

Evaluate:

$\left ( f\circ g\right ) (40)$

Possible Answers:

Correct answer:

Explanation:

$\left ( f\circ g\right ) (x) = f(g(x)) = f (\sqrt{x+9})$

Substitute $\sqrt{x+9}$ for in the definition of :

$f (x) = x^{2} - 10$

$f (\sqrt{x+9} ) =\left (\sqrt{x+9} \right )^{2} - 10$

$f (\sqrt{x+9} ) =x+9- 10$

$f (\sqrt{x+9} ) =x-1$

Therefore,

$\left ( f\circ g\right ) (x) =x-1$

and

$\left ( f\circ g\right ) (40) =40-1 = 39$

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Example Question #831 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Define an operation $\bigstar$ on two real numbers as follows:

For any real numbers a,b ,

$a \bigstar b = a^{2} + b^{2}$

Evaluate: $5 \bigstar 11$

Possible Answers:

1,296

256

146

Correct answer:

146

Explanation:

Substitute a = 5, b = 11 in the definition:

$a \bigstar b = a^{2} + b^{2}$

$5 \bigstar 11 = 5^{2} + 11^{2} = 25 + 121 = 146$

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All ISEE Upper Level Math Resources

6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

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ISEE Upper Level Math : Equations

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #121 : Equations

Example Question #122 : Equations

Example Question #123 : Equations

Example Question #121 : Equations

Example Question #122 : Equations

Example Question #122 : Equations

Example Question #123 : Equations

Example Question #122 : Algebraic Concepts

Example Question #122 : Algebraic Concepts

Example Question #831 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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