How to find a rational number from an exponent

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SAT Math › How to find a rational number from an exponent

Questions 1 - 6

1

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

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

$\frac{\sqrt[3]{81}}{\sqrt[3]{3}} = \frac{3}{1}= 3$

$\frac{81}{3}= 27$

$\sqrt[3]{81\star 3}= \sqrt[3]{243} = 6.24$

$\frac{\sqrt[3]{9\star 9}}{\sqrt[3]{9}}= \sqrt[3]{9}$

Explanation

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

2

Solve for .

$\log _{5}x=2$

Explanation

$log_5{x} = 2$

$x = 5^{2} = 25$

3

Rationalize the denominator:

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

$2+\sqrt{3}$

$2-\sqrt{3}$

$\frac{2+\sqrt{3}}{7}$

$\sqrt{3}$

Explanation

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

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

4

Solve for :

$3^{x-5} = \frac{1}{27}$

Explanation

$3^{x-5}=\frac{1}{27}=3^{-3}$

5

If,

$log_a{120}= log_b{120}$

What does

$\frac{a}{b}= 120$

Explanation

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

then .

6

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

Explanation

From the equation in the problem statement

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

Now squaring both sides we get

$x+3=x^{2}+2x+1$ this is a quadratic equation which equals

$x^{2}+x -2 = 0$

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.

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