Law of multiplication of indices

a^m x aⁿ = a^{m + m}

Law of division of indices

a^m ÷ aⁿ = $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{-\frac{1}{4}+\frac{1}{2}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{\mathrm{-1\; x\; 1\; +\; 2\; x\; 1}}{4}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{\mathrm{-1\; +\; 2}}{4}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{1}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = $\frac{{\mathrm{a}}^{\frac{1}{4}}}{{\mathrm{a}}^{-\frac{1}{2}}}$

Apply a^m ÷ aⁿ = $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{1}{4}-\; -\frac{1}{2}}$

Note: -- = +

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{1}{4}+\frac{1}{2}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{\mathrm{1\; x\; 1\; +\; 2\; x\; 1}}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{\mathrm{1\; +\; 2}}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{3}{4}}$

4^x = $\frac{1}{1024}$

Express both the left and the righ sides to be in the same base.

1024 = 4 x 4 x 4 x 4 x 4 = 4⁵

4^x = $\frac{1}{4^5}$

From indices,$\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$ = a^-b

4^x = 4^-5

Since the bases are the same, the powers are also the same.

x = -5

9 = 3 x 3 = 3²
27 = 3 x 3 x 3 = 3³

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

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

3^{2(y + 1)} = 3^{-3(y - 2)}

Since the bases are the same, the powers are also the same.

2(y + 1) = -3(y - 2)
2y + 2 = -3y + 6

Note: - x - = +
-3 x -2 = + 6

2y + 2 = -3y + 6
2y + 3y = 6 - 2
5y = 4

Divide both sides by 5

y = $\frac{4}{5}$

Before you can simplify a power in an equation, ensure that both the left and the right have the same base.

Express the 32 to base 2.

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

Solve: $2^{\sqrt{\mathrm{2x\; +\; 1}}}$ = 2⁵

Since the bases are the same, you can equate the powers.

$\sqrt{\mathrm{2x\; +\; 1}}$ = 5

Get rid of the square root by squaring both sides of the equation.

2x + 1 = 5²

2x + 1 = 25

2x = 25 - 1

2x = 24

Divide both sides by 2.

x = $\frac{24}{2}$ = 12

To solve for the x, make sure each of the numbers has the same base in order for you to equate the powers.

The lowest base is 2, hence express all the numbers to base 2.

16 = 2 x 2 x 2 x 2 = 2⁴

4 = 2 x 2 = 2²

8 = 2 x 2 x 2 = 2³

16 x 2^{(x + 1)} = 4^x x 8^{(1 − x)}

2⁴ x 2^{(x + 1)} = 2^{2 x x} x 2^{3 x (1 − x)}

2⁴ x 2^{(x + 1)} = 2^2x x 2^{3(1 − x)}

Apply the law of multiplication of indices, which states that when indices of the same base are multiplying, you can add the powers.

2^{4 + x + 1} = 2^{2x + 3(1 - x)}

Since the base at the left equals to the base at the right, you can equate the powers.

4 + x + 1 = 2x + 3(1 - x)

Expand the bracket.

x + 5 = 2x + 3 - 3x

x + 5 = 3 - x

x + x = 3 - 5

2x = -2

Divide both sides by 2.

x = $\frac{-2}{2}$ = -1

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:

1. 16 = 4 x 4 = 4²

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

2. 9 = 3 x 3 = 3²

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

3. 16 = 2 x 2 x 2 x 2 = 2⁴

Expressed this way in order to cancel the denominator 4 of the $\frac{-3}{4}$ power

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{4^2}{3^2}\right)}^{\frac{-3}{2}}x{\left(2^4\right)}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

From the law of indices,

${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{c}}$ = $\frac{{\mathrm{a}}^{\mathrm{c}}}{{\mathrm{b}}^{\mathrm{c}}}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{4^{\mathrm{2\; x}\frac{-3}{2}}}{3^{\mathrm{2\; x}\frac{-3}{2}}}x{2}^{\mathrm{4\; x}\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:
1. The power 2 cancel the denominator 2 of the power $\frac{-3}{2}$

2. The power 4 cancel the denominator of 4 of the $\frac{-3}{4}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{4^{-3}}{3^{-3}}x{2}^{-3}]}^{\frac{1}{3}}$

$\frac{4^{-3}}{3^{-3}}$ = ${\left(\frac{4}{3}\right)}^{-3}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{4}{3}\right)}^{-3}x{2}^{-3}]}^{\frac{1}{3}}$

The $\frac{1}{3}$ affects each of the powers in the bracket.

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${{\left(\frac{4}{3}\right)}^{-3x\frac{1}{3}}x{2}^{-3x\frac{1}{3}}}^{}$

The power 3 cancels the denominator 3 in $\frac{1}{3}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${{\left(\frac{4}{3}\right)}^{-1}x{2}^{-1}}^{}$

Notes:

1. The power -1 means inverse which means you reciprocate (the numerator becomes the denominator and the denominator becomes the numerator).

2. ${\left(\frac{4}{3}\right)}^{-1}$ = $\frac{3}{4}$

3. Every number is being divided by 1 but it is not written.

2 = $\frac{2}{1}$

4. 2^-1 = $\frac{1}{2}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{3}{4}$ x $\frac{1}{2}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{\mathrm{3\; x\; 1}}{\mathrm{4\; x\; 2}}$ = $\frac{3}{8}$

2(x² - y³)

When x = 3 and y = -1

2(x² - y³) = 2 x (3² - (-1)³)

(-1)³ = - 1 x -1 x -1 = 1 x -1 = -1

2(x² - y³) = 2 x (9 - (-1))

2(x² - y³) = 2 x (9 - -1)

Note: - - = + or -(-) = +

2(x² - y³) = 2 x (9 + 1)

2(x² - y³) = 2 x (10)

2(x² - y³) = 2 x 10

2(x² - y³) = 20

Express the decimal in the form a x 10^b where a is a whole number

0.064 = 0.⌒0⌒6⌒4 = 64 x 10^-3

Note: the power is negative since the decimal is moved to the right.

${\left(0.064\right)}^{-\frac{1}{3}}$ = ${\left(\mathrm{64\; x}{10}^{-3}\right)}^{-\frac{1}{3}}$

64 = 4³

${\left(0.064\right)}^{-\frac{1}{3}}$ = ${(4^{3}x{10}^{-3})}^{-\frac{1}{3}}$

The power outside affects each of the powers in the bracket.

${\left(0.064\right)}^{-\frac{1}{3}}$ = $4^{\mathrm{3\; x\; -}\frac{1}{3}}x{10}^{\mathrm{-3\; x\; -}\frac{1}{3}}$

Notes:

1. The 3 cancel each other.

2. -1 x -1 = 1

${\left(0.064\right)}^{-\frac{1}{3}}$ = 4^-1 x 10¹

Note:

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

4^-1 = $\frac{1}{4^1}$ = $\frac{1}{4}$

${\left(0.064\right)}^{-\frac{1}{3}}$ = $\frac{1}{4}$ x 10 = $\frac{5}{2}$

a² - b² = (a + b)(a - b) known as difference of two squares.

(27.63)² - (12.37)² = (27.63 + 12.37)(27.63 - 12.37)

(27.63)² - (12.37)² = (40)(15.26)

(27.63)² - (12.37)² = 610.4 ≈ 610 (three significant figures)

Note: 0 in the middle or end of a number is significant.

0.25 = $\frac{25}{100}$ = $\frac{1}{4}$ = $\frac{1}{2^2}$

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

$\frac{1}{2^2}$ = 2^-2

(0.25)^y = (2^-2)^y

(0.25)^y = 2^{-2 x y}

(0.25)^y = 2^-2y

2^-2y = 32

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

2^-2y = 2⁵

Since the bases are the same, the powers are also the same.

-2y = 5

Divide both sides by -2

y = $\frac{5}{-2}$ = -$\frac{5}{2}$

$\sqrt{a}$ = $a^{\frac{1}{2}}$

$\sqrt{\mathrm{64}}$ = ${\mathrm{64}}^{\frac{1}{2}}$

64 = 2⁶

$\sqrt{\mathrm{64}}$ = $2^{\mathrm{6\; x}\frac{1}{2}}$ = 2³

2a = 2³

Since the bases are the same, the powers are also the same.

a = 3

$\frac{\mathrm{b}}{\mathrm{a}}$ = 3

b = 3 x a

b = 3 x 3

b = 9

a² + b² = 3² + 9²

a² + b² = 9 + 81

a² + b² = 90

3 x 9 ^{1 + x} = 27^-x

Concept

A^b x A^c = A^b+c

Thus if the bases are the same, we can add the indexes (superscripts)

3 can go into the other numbers(9,27) hence we change then to have base 3

9 = 3¹x 3¹ = 3^{1 + 1} = 3²

27 = 3¹x 3¹ x 3¹ =3¹⁺¹⁺¹ = 3³

⇒ 3 x 3^{2(1 + x)} = 3^3(-x)

But 3 = 3¹, thus every number is raised to the power 1
⇒ 3¹ x 3^{2(1 + x)} = 3^3(-x)

Applying A^b x A^c = A^{b + c}

3 ¹ x 3^{2(1+ x)} = 3^{1 + 2(1 + x )}

3^{1+2(1+ x)} = 3^3(-x)

Concept
If A^b = A^c ⇒ b = c

Thus if the base at the left side of the equation, is the same as the right side of the equation, then the powers are the same

∴ 1 + 2(1 + x ) = 3(-x)

Concept

a(b + c ) = a x b + a x c

a(b - c ) = a x b - a x c

Thus the number outside the bracket multiplies each of the number in the bracket and the sign ( + or - ) is(are) maintained

-x = -1 x x

3(-x) = 3 x (-1) x x = -3x

1 + 2(1 + x ) = 1 + 2 x 1 + 2 x x

1 + 2(1 + x ) = 1 + 2 + 2x

1 + 2(1 + x ) = 3 + 2x

1 + 2(1+x) = 3(-x)

⇒ 3 + 2x = -3x

Grouping the x at the left and the rest at the right

Concept
If the sign is positive(+) and crosses to the other side of the equation the sign changes to negative ( - ) and the vice versa

2x + 3x = -3

5x = -3

Concept
Divide both sides by 5

x = $\frac{-3}{5}$ = -$\frac{3}{5}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$

Notes:

1. a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

2. a^-1 = $\frac{1}{{\mathrm{a}}^1}$ = $\frac{1}{\mathrm{a}}$

3. x^-1 = $\frac{1}{{\mathrm{x}}^1}$ = $\frac{1}{\mathrm{x}}$

4. y^-1 = $\frac{1}{{\mathrm{y}}^1}$ = $\frac{1}{\mathrm{y}}$

x^-1 + y^-1 = $\frac{1}{\mathrm{x}}$ + $\frac{1}{\mathrm{y}}$

x^-1 + y^-1 = $\frac{\mathrm{1\; x\; y\; +\; 1\; x\; x}}{\mathrm{xy}}$

x^-1 + y^-1 = $\frac{\mathrm{y\; +\; x}}{\mathrm{xy}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$ ÷ (x + y)

Every number or expression is being divided by 1 but not written

x + y = $\frac{\mathrm{x\; +\; y}}{1}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$ ÷ $\frac{\mathrm{x\; +\; y}}{1}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$ x $\frac{1}{\mathrm{x\; +\; y}}$

x + y cancels each other.

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{1}{\mathrm{xy}}$ x $\frac{1}{1}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{1}{\mathrm{xy}}$

3^-x = k

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

3^-x = $\frac{1}{3^{\mathrm{x}}}$

$\frac{1}{3^{\mathrm{x}}}$ = k

Make 3^x the subject.

Get rid of the fraction by multiplying both sides by the denominator, 3^x

3^x x $\frac{1}{3^{\mathrm{x}}}$ = k x 3^x

1 = k x 3^x

Divide both sides by k.

$\frac{1}{\mathrm{k}}$ = 3^x

3^x = $\frac{1}{\mathrm{k}}$

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

Law of division of indices

a^m ÷ aⁿ = $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{-\frac{1}{4}+\frac{1}{2}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{\mathrm{-1\; x\; 1\; +\; 2\; x\; 1}}{4}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{\mathrm{-1\; +\; 2}}{4}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{1}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = $\frac{{\mathrm{a}}^{\frac{1}{4}}}{{\mathrm{a}}^{-\frac{1}{2}}}$

Apply a^m ÷ aⁿ = $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{1}{4}-\; -\frac{1}{2}}$

Note: -- = +

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{1}{4}+\frac{1}{2}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{\mathrm{1\; x\; 1\; +\; 2\; x\; 1}}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{\mathrm{1\; +\; 2}}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{3}{4}}$

4^x = $\frac{1}{1024}$

Express both the left and the righ sides to be in the same base.

1024 = 4 x 4 x 4 x 4 x 4 = 4⁵

4^x = $\frac{1}{4^5}$

From indices,$\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$ = a^-b

4^x = 4^-5

Since the bases are the same, the powers are also the same.

x = -5

9 = 3 x 3 = 3²
27 = 3 x 3 x 3 = 3³

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

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

3^{2(y + 1)} = 3^{-3(y - 2)}

Since the bases are the same, the powers are also the same.

2(y + 1) = -3(y - 2)
2y + 2 = -3y + 6

Note: - x - = +
-3 x -2 = + 6

2y + 2 = -3y + 6
2y + 3y = 6 - 2
5y = 4

Divide both sides by 5

y = $\frac{4}{5}$

Before you can simplify a power in an equation, ensure that both the left and the right have the same base.

Express the 32 to base 2.

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

Solve: $2^{\sqrt{\mathrm{2x\; +\; 1}}}$ = 2⁵

Since the bases are the same, you can equate the powers.

$\sqrt{\mathrm{2x\; +\; 1}}$ = 5

Get rid of the square root by squaring both sides of the equation.

2x + 1 = 5²

2x + 1 = 25

2x = 25 - 1

2x = 24

Divide both sides by 2.

x = $\frac{24}{2}$ = 12

To solve for the x, make sure each of the numbers has the same base in order for you to equate the powers.

The lowest base is 2, hence express all the numbers to base 2.

16 = 2 x 2 x 2 x 2 = 2⁴

4 = 2 x 2 = 2²

8 = 2 x 2 x 2 = 2³

16 x 2^{(x + 1)} = 4^x x 8^{(1 − x)}

2⁴ x 2^{(x + 1)} = 2^{2 x x} x 2^{3 x (1 − x)}

2⁴ x 2^{(x + 1)} = 2^2x x 2^{3(1 − x)}

Apply the law of multiplication of indices, which states that when indices of the same base are multiplying, you can add the powers.

2^{4 + x + 1} = 2^{2x + 3(1 - x)}

Since the base at the left equals to the base at the right, you can equate the powers.

4 + x + 1 = 2x + 3(1 - x)

Expand the bracket.

x + 5 = 2x + 3 - 3x

x + 5 = 3 - x

x + x = 3 - 5

2x = -2

Divide both sides by 2.

x = $\frac{-2}{2}$ = -1

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:

1. 16 = 4 x 4 = 4²

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

2. 9 = 3 x 3 = 3²

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

3. 16 = 2 x 2 x 2 x 2 = 2⁴

Expressed this way in order to cancel the denominator 4 of the $\frac{-3}{4}$ power

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{4^2}{3^2}\right)}^{\frac{-3}{2}}x{\left(2^4\right)}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

From the law of indices,

${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{c}}$ = $\frac{{\mathrm{a}}^{\mathrm{c}}}{{\mathrm{b}}^{\mathrm{c}}}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{4^{\mathrm{2\; x}\frac{-3}{2}}}{3^{\mathrm{2\; x}\frac{-3}{2}}}x{2}^{\mathrm{4\; x}\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:
1. The power 2 cancel the denominator 2 of the power $\frac{-3}{2}$

2. The power 4 cancel the denominator of 4 of the $\frac{-3}{4}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{4^{-3}}{3^{-3}}x{2}^{-3}]}^{\frac{1}{3}}$

$\frac{4^{-3}}{3^{-3}}$ = ${\left(\frac{4}{3}\right)}^{-3}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{4}{3}\right)}^{-3}x{2}^{-3}]}^{\frac{1}{3}}$

The $\frac{1}{3}$ affects each of the powers in the bracket.

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${{\left(\frac{4}{3}\right)}^{-3x\frac{1}{3}}x{2}^{-3x\frac{1}{3}}}^{}$

The power 3 cancels the denominator 3 in $\frac{1}{3}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${{\left(\frac{4}{3}\right)}^{-1}x{2}^{-1}}^{}$

Notes:

1. The power -1 means inverse which means you reciprocate (the numerator becomes the denominator and the denominator becomes the numerator).

2. ${\left(\frac{4}{3}\right)}^{-1}$ = $\frac{3}{4}$

3. Every number is being divided by 1 but it is not written.

2 = $\frac{2}{1}$

4. 2^-1 = $\frac{1}{2}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{3}{4}$ x $\frac{1}{2}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{\mathrm{3\; x\; 1}}{\mathrm{4\; x\; 2}}$ = $\frac{3}{8}$

0.25 = $\frac{25}{100}$ = $\frac{1}{4}$ = $\frac{1}{2^2}$

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

$\frac{1}{2^2}$ = 2^-2

(0.25)^y = (2^-2)^y

(0.25)^y = 2^{-2 x y}

(0.25)^y = 2^-2y

2^-2y = 32

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

2^-2y = 2⁵

Since the bases are the same, the powers are also the same.

-2y = 5

Divide both sides by -2

y = $\frac{5}{-2}$ = -$\frac{5}{2}$

$\sqrt{a}$ = $a^{\frac{1}{2}}$

$\sqrt{\mathrm{64}}$ = ${\mathrm{64}}^{\frac{1}{2}}$

64 = 2⁶

$\sqrt{\mathrm{64}}$ = $2^{\mathrm{6\; x}\frac{1}{2}}$ = 2³

2a = 2³

Since the bases are the same, the powers are also the same.

a = 3

$\frac{\mathrm{b}}{\mathrm{a}}$ = 3

b = 3 x a

b = 3 x 3

b = 9

a² + b² = 3² + 9²

a² + b² = 9 + 81

a² + b² = 90

3 x 9 ^{1 + x} = 27^-x

Concept

A^b x A^c = A^b+c

Thus if the bases are the same, we can add the indexes (superscripts)

3 can go into the other numbers(9,27) hence we change then to have base 3

9 = 3¹x 3¹ = 3^{1 + 1} = 3²

27 = 3¹x 3¹ x 3¹ =3¹⁺¹⁺¹ = 3³

⇒ 3 x 3^{2(1 + x)} = 3^3(-x)

But 3 = 3¹, thus every number is raised to the power 1
⇒ 3¹ x 3^{2(1 + x)} = 3^3(-x)

Applying A^b x A^c = A^{b + c}

3 ¹ x 3^{2(1+ x)} = 3^{1 + 2(1 + x )}

3^{1+2(1+ x)} = 3^3(-x)

Concept
If A^b = A^c ⇒ b = c

Thus if the base at the left side of the equation, is the same as the right side of the equation, then the powers are the same

∴ 1 + 2(1 + x ) = 3(-x)

Concept

a(b + c ) = a x b + a x c

a(b - c ) = a x b - a x c

Thus the number outside the bracket multiplies each of the number in the bracket and the sign ( + or - ) is(are) maintained

-x = -1 x x

3(-x) = 3 x (-1) x x = -3x

1 + 2(1 + x ) = 1 + 2 x 1 + 2 x x

1 + 2(1 + x ) = 1 + 2 + 2x

1 + 2(1 + x ) = 3 + 2x

1 + 2(1+x) = 3(-x)

⇒ 3 + 2x = -3x

Grouping the x at the left and the rest at the right

Concept
If the sign is positive(+) and crosses to the other side of the equation the sign changes to negative ( - ) and the vice versa

2x + 3x = -3

5x = -3

Concept
Divide both sides by 5

x = $\frac{-3}{5}$ = -$\frac{3}{5}$

3^-x = k

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

3^-x = $\frac{1}{3^{\mathrm{x}}}$

$\frac{1}{3^{\mathrm{x}}}$ = k

Make 3^x the subject.

Get rid of the fraction by multiplying both sides by the denominator, 3^x

3^x x $\frac{1}{3^{\mathrm{x}}}$ = k x 3^x

1 = k x 3^x

Divide both sides by k.

$\frac{1}{\mathrm{k}}$ = 3^x

3^x = $\frac{1}{\mathrm{k}}$

Solve the inequality.

$\frac{1}{3}$(2x - 9) ≥ 2(2x + 1)

Get rid of th fraction by multiplying both sides of the equation by the denominator, 3

3 x $\frac{1}{3}$(2x - 9) ≥ 3 x 2(2x + 1)

1 x (2x - 9) ≥ 6(2x + 1)

2x - 9 ≥ 12x + 6

2x - 12x ≥ 6 + 9

-10x ≥ 15

Divide both sides by -10. When dividing by a negative, the inequality sign changes.

x ≤ $\frac{15}{-10}$

x ≤ -1.5

Notes:

1. -1.5 is between -1 and - 2

2. For ≥ and ≤, the solid is shaded

3. For ≤, the arrow is to the left just as the direction of the sign and for ≥, the arrow is to the right