Exponential Ratios and Rational Numbers - PSAT Math

Card 0 of 133

Which of the following lists the above quantities from least to greatest?

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If a piece of pie is cut into 3 sections, and each of those pieces is further cut into three sections, then those pieces are cut into three sections, how many (tiny) pieces of pie are there?

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The answer is 33 = 27

If and are positive integers and , then what is the value of $\frac{m}{n}$ ?

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43 = 64

Alternatively written, this is 4(4)(4) = 64 or 43 = 641.

Thus, m = 3 and n = 1.

m/n = 3/1 = 3.

Write the following logarithm in expanded form:

$\log $x^{2}$y$

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$\log $x^{2}$y=\log $x^{2}$+\log y=2\log x+\log y$

$Simplify: $\frac{\sqrt[3]{81}$}{\sqrt[3]{3}}$

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$$\frac{\sqrt[3]{81}$}{\sqrt[3]{3}} = \sqrt[3]{$\frac{81}{3}$}= \sqrt[3]{27}= 3$

$$\sqrt{x+3}$ -x = 1$

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From the equation in the problem statement

$$\sqrt{x+3}$ = x + 1$

Now squaring both sides we get

this is a quadratic equation which equals

and the factors of this equation are

$\left ( x+2 \right )\left ( x-1 \right )$

This gives us $x=-2\ or\ x=1$ .

However, if we plug these solutions back into the original equation, does not create an equality. Therefore, is an extraneous solution.

Rationalize the denominator:

$$\frac{1}{2-\sqrt{3}$}$

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The conjugate of $2-$\sqrt{3}$$ is $2+$\sqrt{3}$$ .

Now multiply both the numerator and the denominator by $2+$\sqrt{3}$$

and you get:

$$\frac{2+\sqrt{3}$}\left ({2-$\sqrt{3}$} \right )\left ( 2+$\sqrt{3}$ \right )$

$\left ( 2-$\sqrt{3}$ \right )\left ( 2+$\sqrt{3}$ \right ) $=2^{2}$-\left ( $\sqrt{3}$ \right $)^{2}$ =4-3 = 1$

Hence we get

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

Solve for :

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Solve for .

$\log _{5}x=2$

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$log_5{x} = 2$

$x = $5^{2}$ = 25$

If,

$log_a{120}= log_b{120}$

What does

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If $log_a{x} = log_b{x}$ ,

then .

For some positive integer , if , what is the value of ?

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If , then must equal 3 (Note that cannot be -3 because you need it to be positive.

Now, plug into the new equation :

Solve for .

$2^{x}$= 64

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Since $2^{x}$= $2^{6}$

Hence

Simplify:

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Solve for :

$$\sqrt{x}$ -1= 4$

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From the equation one can see that

$$\sqrt{x}$ = 5$

Hence must be equal to 25.

Evaluate:

$\log 0.1=$

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$0.1= $10^{-1}$$

$\log 0.1=-1$

Solve for .

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$x= $2^4${}= 16$

Solve for .

$\log _{9}x=\frac{1}{2}$$

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$\log _{9}x=\frac{1}{2}$$

$x = $9^{$\frac{1}${2}$}= 3$

Solve for .

$\log (x+3)+\log x=1$

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Use the rules of logarithms to combine terms.

$\log (x(x+3))=1$

Hence,

By fatoring we get

Hence $x=-5\ or\ 2$ .

However, you cannot take the logarithm of a negative number. Thus, the only value for is .

$\log 3=a$

and

$\log 5=b$

Find $\log 15$

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$\log 15=\log (5\cdot 3)=\log 5+\log 3$

Hence the correct answer is .

Solve for .

$2^{x}$= 64

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Since $2^{x}$= $2^{6}$

Hence