Powers and Exponents Worksheet for practicing simplification, exponential form, and value calculation.

Here are the step-by-step solutions for each problem on the worksheet.

1. Simplify

a) $\left(\frac{2}{3}\right)^9 \times \left(\frac{4}{9}\right)^{-6}$

* Step 1: Notice that $\frac{4}{9}$ is the square of $\frac{2}{3}$ because $2^2 = 4$ and $3^2 = 9$. So, we can rewrite $\frac{4}{9}$ as $\left(\frac{2}{3}\right)^2$.
* Step 2: Substitute this into the expression: $\left(\frac{2}{3}\right)^9 \times \left[\left(\frac{2}{3}\right)^2\right]^{-6}$.
* Step 3: Use the power rule $(x^a)^b = x^{a \cdot b}$. Multiply the exponents for the second part: $2 \times -6 = -12$. Now we have: $\left(\frac{2}{3}\right)^9 \times \left(\frac{2}{3}\right)^{-12}$.
* Step 4: When multiplying terms with the same base, add the exponents: $9 + (-12) = -3$. The expression becomes $\left(\frac{2}{3}\right)^{-3}$.
* Step 5: A negative exponent means taking the reciprocal (flipping the fraction) and making the exponent positive: $\left(\frac{3}{2}\right)^3$.
* Step 6: Calculate the value: $\frac{3^3}{2^3} = \frac{27}{8}$.

b) $\frac{(3b^3)^9}{(9b^2)^4}$

* Step 1: Expand the numerator using the rule $(xy)^n = x^n y^n$: $(3b^3)^9 = 3^9 \cdot (b^3)^9 = 3^9 b^{27}$.
* Step 2: Expand the denominator. Note that $9 = 3^2$. So, $(9b^2)^4 = (3^2 b^2)^4 = (3^2)^4 \cdot (b^2)^4 = 3^8 b^8$.
* Step 3: Divide the numerator by the denominator: $\frac{3^9 b^{27}}{3^8 b^8}$.
* Step 4: Subtract the exponents for like bases ($\frac{x^a}{x^b} = x^{a-b}$).
* For the numbers: $3^{9-8} = 3^1 = 3$.
* For the variables: $b^{27-8} = b^{19}$.
* Result: $3b^{19}$.

c) $\frac{(4x^2)^6}{(2x^4)^4}$

* Step 1: Expand the numerator: $(4x^2)^6 = 4^6 \cdot (x^2)^6 = 4^6 x^{12}$.
* Step 2: Expand the denominator: $(2x^4)^4 = 2^4 \cdot (x^4)^4 = 2^4 x^{16}$.
* Step 3: Simplify the numbers. Since $4 = 2^2$, then $4^6 = (2^2)^6 = 2^{12}$. The fraction of numbers is $\frac{2^{12}}{2^4} = 2^{12-4} = 2^8$. We know that $2^8 = 256$.
* Step 4: Simplify the variables: $\frac{x^{12}}{x^{16}} = x^{12-16} = x^{-4}$. This can be written as $\frac{1}{x^4}$.
* Result: $\frac{256}{x^4}$.

d) $(-4)^{100} \times (-4)^{20}$

* Step 1: When multiplying bases that are the same, add the exponents: $100 + 20 = 120$.
* Step 2: The expression becomes $(-4)^{120}$.
* Step 3: Since the exponent (120) is an even number, the negative sign disappears (a negative number multiplied by itself an even number of times results in a positive number). So, $(-4)^{120} = 4^{120}$.

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2. Find the value of $x$

a) $5^{\left(\frac{2}{5}\right)} = 5^x$

* Step 1: Since the bases are both 5, the exponents must be equal to each other.
* Result: $x = \frac{2}{5}$.

b) $\left(\frac{8}{9}\right)^5 \times \left(\frac{9}{4}\right) = 2x$

* Step 1: First, simplify the left side. $\left(\frac{8}{9}\right)^5 = \frac{32768}{59049}$.
* Step 2: Multiply by $\frac{9}{4}$: $\frac{32768}{59049} \times \frac{9}{4}$.
* Step 3: Simplify the multiplication:
* $32768 \div 4 = 8192$.
* $59049 \div 9 = 6561$.
* So, Left Side $= \frac{8192}{6561}$.
* Step 4: Set equal to $2x$: $\frac{8192}{6561} = 2x$.
* Step 5: Divide by 2 to solve for $x$: $x = \frac{8192}{6561 \times 2} = \frac{4096}{6561}$.

c) $6^x = 216$

* Step 1: Express 216 as a power of 6.
* $6^1 = 6$
* $6^2 = 36$
* $6^3 = 216$
* Step 2: So, $6^x = 6^3$.
* Result: $x = 3$.

d) $x(3^{-5}) = 3$

* Step 1: Isolate $x$ by dividing both sides by $3^{-5}$: $x = \frac{3}{3^{-5}}$.
* Step 2: Use exponent rules for division ($\frac{a^m}{a^n} = a^{m-n}$). Note that $3$ is $3^1$.
* $x = 3^{1 - (-5)} = 3^{1+5} = 3^6$.
* Step 3: Calculate $3^6$: $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729$.
* Result: $x = 729$.

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3. Write in exponential form

a) $\left(\frac{2}{5}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2}$

* Step 1: The base $\left(\frac{2}{5}\right)$ is repeated 3 times.
* Step 2: Add the exponents: $-2 + (-2) + (-2) = -6$.
* Result: $\left(\frac{2}{5}\right)^{-6}$.

b) $\left(\frac{5}{2}\right)^{-1} \times \left(\frac{5}{2}\right)^{-1} \times \left(\frac{5}{2}\right)^{-1} \times \left(\frac{5}{2}\right)^{-1}$

* Step 1: The base $\left(\frac{5}{2}\right)$ is repeated 4 times.
* Step 2: Add the exponents: $-1 + (-1) + (-1) + (-1) = -4$.
* Result: $\left(\frac{5}{2}\right)^{-4}$.

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4. Find the values of the following

a) $(3^{-1} + 4^{-1} + 5^{-1})^0$

* Step 1: Recall the zero exponent rule: Any non-zero number raised to the power of 0 equals 1 ($a^0 = 1$).
* Step 2: Check if the inside is zero. $3^{-1} = \frac{1}{3}$, $4^{-1} = \frac{1}{4}$, $5^{-1} = \frac{1}{5}$. Their sum is clearly not zero.
* Result: $1$.

b) $(3^0 + 4^{-1}) \times 2^2$

* Step 1: Evaluate the terms inside the parenthesis.
* $3^0 = 1$.
* $4^{-1} = \frac{1}{4}$.
* So, $(1 + \frac{1}{4}) = \frac{5}{4}$.
* Step 2: Evaluate the term outside.
* $2^2 = 4$.
* Step 3: Multiply them together: $\frac{5}{4} \times 4$.
* Step 4: The 4s cancel out.
* Result: $5$.

Final Answer:
1. Simplify
a) $\frac{27}{8}$
b) $3b^{19}$
c) $\frac{256}{x^4}$
d) $4^{120}$

2. Find the value of x
a) $\frac{2}{5}$
b) $\frac{4096}{6561}$
c) $3$
d) $729$

3. Write in exponential form
a) $\left(\frac{2}{5}\right)^{-6}$
b) $\left(\frac{5}{2}\right)^{-4}$

4. Find the values
a) $1$
b) $5$