When dealing with fractional exponents, we write as 

$\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}$ 

which $\displaystyle b$ is the index of the radical, $\displaystyle a$ is the exponent raising base $\displaystyle x$. 

We evaluate negative exponents as 

$\displaystyle x^{-a}=\frac{1}{x^a}$ 

which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

Therefore 

$\displaystyle \frac{1}{4}^{-\frac{3}{4}}=\frac{1}{\frac{1}{\sqrt[4]{4^3}}}=\sqrt[4]{64}$.

We can factor out $\displaystyle 2^4$ or $\displaystyle 16$. 

$\displaystyle \sqrt[4]{64}=\sqrt[4]{16}*\sqrt[4]{4}=2*\sqrt[4]{4}=2\sqrt[4]{4}$

In order to simplify this, we will need to multiply both the numerator and denominator with the outside power according to the power rule of exponents.

$\displaystyle (\frac{3}{4})^{\frac{4}{3}} = \frac{3^\frac{4}{3}}{4^\frac{4}{3}}$

The numerator of the power represents the power raised to.  The denominator of the power represents the root of the radical.  Rewrite the fractions.

$\displaystyle \frac{3^\frac{4}{3}}{4^\frac{4}{3}} = \frac{(\sqrt[3]{3})^4}{(\sqrt[3]{4})^4} = \frac{\sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}}{\sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}}$

Recall that multiplying the radicals in the third root three times will leave the integer by itself.

The expression becomes:  

$\displaystyle \frac{\sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}}{\sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}}= \frac{3 \sqrt[3]{3}}{4 \sqrt[3]{4}}$

Rationalize the denominator by multiplying the top and bottom by $\displaystyle \sqrt[3]{4}$ twice in order to eliminate the cube root radical denominator.

$\displaystyle \frac{3 \sqrt[3]{3}}{4 \sqrt[3]{4}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}}= \frac{3 \cdot \sqrt[3]{48}}{4(4)} = \frac{3 \sqrt[3]{48}}{16}$

Simplify $\displaystyle \sqrt[3]{48}$.

$\displaystyle \sqrt[3]{48} = \sqrt[3]{8\cdot 6} = \sqrt[3]{8} \cdot \sqrt[3]{6} = 2 \sqrt[3]{6}$

Replace the new term and reduce the fraction.  The fraction becomes:

$\displaystyle \frac{3 \sqrt[3]{48}}{16} = \frac{3 \cdot 2 \sqrt[3]{6}}{16} = \frac{3\sqrt[3]{6}}{8}$

The answer is:  $\displaystyle \frac{3\sqrt[3]{6}}{8}$

Use the product rule of exponents to simplify this term.

$\displaystyle (\frac{2}{3})^{\frac{5}{2}} = (\frac{2^{\frac{5}{2}}}{3^{\frac{5}{2}}})$

Rewrite this using radicals.  The numerator represents the power that the radical is raised to.  The denominator represents the root.

$\displaystyle (\frac{2^{\frac{5}{2}}}{3^{\frac{5}{2}}})= \frac{(\sqrt{2})^5}{(\sqrt{3})^5} = \frac{\sqrt{2}\cdot \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}}{\sqrt{3}\cdot\sqrt{3}\cdot\sqrt{3}\cdot\sqrt{3}\cdot\sqrt{3}}$

Multiply the terms together.  A radical multiplied by itself will be the integer inside the radical.  The terms become:

$\displaystyle \frac{2\cdot 2 \cdot\sqrt2}{3\cdot3\cdot\sqrt3} = \frac{4\sqrt2}{9\sqrt{3}}$

Rationalize the denominator.  Multiply the top and bottom by square root three.

$\displaystyle \frac{4\sqrt2}{9\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt6}{9\times 3}$

The answer is:  $\displaystyle \frac{4\sqrt6}{27}$

The fractional exponent will include both the power and the root.  The numerator will represent the power that the quantity is raised to, and the denominator represents the root of the term.

Rewrite the expression in radical form.

$\displaystyle (3)^{\frac{7}{9}} =( \sqrt[9]{3})^7$

The answer is:  $\displaystyle ( \sqrt[9]{3})^7$

Rewrite the half power with a radical.

$\displaystyle (\frac{1}{3})^{\frac{1}{2}} = \sqrt{\frac{1}{3}}$

Split the radical as two radicals.

$\displaystyle \sqrt{\frac{1}{3}} = \frac{\sqrt1}{\sqrt3}$

Rationalize the denominator.  Multiply the top and bottom by square root three.

$\displaystyle \frac{\sqrt1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3} =\frac{\sqrt3}{3}$

The answer is:  $\displaystyle \frac{\sqrt3}{3}$

In order to solve this, we will need to rewrite the inner term as a radical.

$\displaystyle [-16^{\frac{3}{2}}]^{\frac{1}{2}} = [-(\sqrt{16})^3] ^{\frac{1}{2}}$

Simplify the inner term.

$\displaystyle [-64]^{\frac{1}{2}} = \sqrt{-64}$

The answer is:  $\displaystyle 8i$

We can rewrite both terms using the radicals.  The denominator of a fractional exponent is the index of the root.  The numerator of the fraction is the power of the quantity.

Rewrite the terms.

$\displaystyle 289^{\frac{3}{2}}-27^{\frac{2}{3}} =( \sqrt{289})^3 - (\sqrt[3]{27})^2$

Simplify the radicals and solve.

$\displaystyle 289^{\frac{3}{2}}-27^{\frac{2}{3}} =(17)^3 - (3)^2 = 4913 -9 = 4904$

The answer is:  $\displaystyle 4904$

Start by simplifying the numerator. Since two terms with the same base are being multiplied, add the exponents.

$\displaystyle \frac{(x^2)(x^{\frac{1}{2}}y^9)}{x^{-\frac{4}{3}}y^5}=\frac{x^{\frac{5}{2}}y^9}{x^{-\frac{4}{3}}y^5}$

Now, when terms with the same bases are divided, subtract the exponent from the denominator from the exponent in the numerator.

The exponent for $\displaystyle x$ is

$\displaystyle \frac{5}{2}-(-\frac{4}{3})=\frac{23}{6}$

The exponent for $\displaystyle y$ is

$\displaystyle 9-5=4$

So then,

$\displaystyle \frac{(x^2)(x^{\frac{1}{2}}y^9)}{x^{-\frac{4}{3}}y^5}=\frac{x^{\frac{5}{2}}y^9}{x^{-\frac{4}{3}}y^5}=x^{\frac{23}{6}}y^4$

1. Use $\displaystyle x=Pe^{rt}$ where $\displaystyle x$ is the current amount, $\displaystyle r$ is the interest rate, $\displaystyle t$ is the amount of time in years since the initial deposit, and $\displaystyle P$ is the amount initially deposited.

$\displaystyle r=0.075$

$\displaystyle P=$10,000$

$\displaystyle t=12$

2. Solve for $\displaystyle x$

$\displaystyle x=Pe^{rt}$

$\displaystyle x=10,000e^{(0.075\cdot12)}$

$\displaystyle x=24,596.03$

You currently have $24,596 in your account.

Step 1: Achieve same bases

$\displaystyle 5^{x}=125^{3x-1}$

$\displaystyle 5^{x}=(5^{3})^{3x-1}$

$\displaystyle 5^{x}=5^{9x-3}$

Step 2: Drop bases, set exponents equal to eachother

$\displaystyle x=9x-3$

Step 3: Solve for x

$\displaystyle -8x=-3$

$\displaystyle x=\frac{3}{8}$