Unlocking the Power of 10: An IB Math Workout on Exponents and Scientific Notation
What’s up, fellow math explorers! Get ready to tackle some of the biggest (and smallest!) numbers imaginable. From the mass of the sun to the width of a single atom, writing out all those zeros is a non-starter. That’s where the elegance of scientific notation and the power of exponents come to the rescue. In this blog post, we’re going to roll up our sleeves and work through some classic IB-style problems. We’ll kick things off by making sure we’re solid on the fundamental Laws of Exponents. Then, we’ll jump into the main event: fluently converting numbers between standard and scientific notation. And because we love a good challenge, we’ll also tackle those slightly sneaky problems where a number is given in a form like 334562.334 x 10^3 and needs to be wrangled into proper scientific notation. This is a core skill for your exams, so let’s get these reps in and build that confidence.
The Warm-Up: Nailing the Fundamentals
Let’s begin with some straightforward questions to get the brain juices flowing. Focus on showing your work and being precise!
Question 1: Hey team! Let’s start with a classic exponent workout. Your mission, should you choose to accept it, is to simplify this expression completely. Make sure your final answer only has positive exponents. Go for it! $$ \frac{(3a^4b^{-3})^2}{18a^{-2}b^5} $$
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Solution: Alright, let’s break this beast down step-by-step. No sweat!
First, tackle the numerator. We need to apply that square to everything inside the parentheses. Remember the ‘power of a power’ rule means we multiply exponents.
$$ (3a^4b^{-3})^2 = 3^2 \cdot (a^4)^2 \cdot (b^{-3})^2 = 9a^{8}b^{-6} $$
Now, let’s pop that back into our fraction:
$$ \frac{9a^{8}b^{-6}}{18a^{-2}b^5} $$
Next up, let’s handle the coefficients (the numbers out front) and then the variables one by one. For division, we subtract exponents.
Coefficients: $ \frac{9}{18} = \frac{1}{2} $
The ‘a’ terms: $ \frac{a^8}{a^{-2}} = a^{8 – (-2)} = a^{10} $
The ‘b’ terms: $ \frac{b^{-6}}{b^5} = b^{-6 – 5} = b^{-11} $
Putting it all together, we get: $ \frac{1}{2}a^{10}b^{-11} $. But wait, the question asked for only positive exponents!
Tip
Remember that a negative exponent just means “move it!” A term with a negative exponent in the numerator moves to the denominator to become positive, and vice-versa. So, $x^{-n} = \frac{1}{x^n}$.
To make $b^{-11}$ positive, we just move it to the denominator. So our final, beautiful answer is:
$$ \frac{a^{10}}{2b^{11}} $$
Question 2: Time for some fractional exponents! These can look a little weird, but they’re just roots in disguise. Try to evaluate the following expression without using a calculator. Show us your mental math muscles! $$ (125^{2/3}) \cdot (16^{-3/4}) $$
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Solution: You got this! The key is to remember what a fractional exponent like $x^{m/n}$ actually means: it’s the $n$-th root of $x$, all raised to the power of $m$. Or, $(\sqrt[n]{x})^m$. It’s almost always easier to do the root part first!
Let’s look at the first term, $125^{2/3}$:
$$ 125^{2/3} = (\sqrt[3]{125})^2 $$
The cube root of 125 is 5 (since $5 \times 5 \times 5 = 125$). So, we have:
$$ (5)^2 = 25 $$
Easy enough! Now for the second term, $16^{-3/4}$. The negative exponent means we’re going to flip it into a fraction.
$$ 16^{-3/4} = \frac{1}{16^{3/4}} = \frac{1}{(\sqrt[4]{16})^3} $$
The fourth root of 16 is 2 (since $2 \times 2 \times 2 \times 2 = 16$). So, we get:
$$ \frac{1}{(2)^3} = \frac{1}{8} $$
Finally, we just multiply our two results together:
$$ 25 \cdot \frac{1}{8} = \frac{25}{8} $$
And there you have it! Not so scary after all, right?
Question 3: Let’s take these skills into the real world (or at least, the microscopic world). The estimated mass of a single water molecule is about $3.0 \times 10^{-23}$ grams. If you have a bottle containing $2.5 \times 10^{24}$ water molecules, what is the total mass of the water in grams? Give your answer in proper scientific notation.
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Solution: This is a perfect job for scientific notation. To find the total mass, we just need to multiply the mass of one molecule by the number of molecules.
$$ \text{Total Mass} = (3.0 \times 10^{-23}) \times (2.5 \times 10^{24}) $$
The strategy is to multiply the numbers and the powers of 10 separately.
Numbers first: $ 3.0 \times 2.5 = 7.5 $
Now the powers of 10. When we multiply, we add the exponents:
$$ 10^{-23} \times 10^{24} = 10^{-23 + 24} = 10^1 $$
Combine them back together:
$$ 7.5 \times 10^1 $$
This is already in proper scientific notation because 7.5 is between 1 and 10. We can also write it as a standard number, since $10^1$ is just 10.
$$ 7.5 \times 10 = 75 \text{ grams} $$
The question asks for the answer in scientific notation, so $7.5 \times 10^1$ grams is the perfect final answer. Awesome work!
Question 4: The Sun’s diameter is approximately $1.4 \times 10^6$ km. A smaller star, Proxima Centauri, has a diameter of about $2.0 \times 10^5$ km. What is the difference between their diameters? Write your final answer in scientific notation.
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Solution: To find the difference, we need to subtract the smaller diameter from the larger one.
$$ \text{Difference} = (1.4 \times 10^6) – (2.0 \times 10^5) $$
Tip
Heads up! You can’t add or subtract numbers in scientific notation unless their powers of 10 are the same. You have to adjust one of them first.
Let’s make the exponents match. It’s usually easiest to change the smaller exponent to match the larger one. We want to turn $10^5$ into $10^6$.
To go from $10^5$ to $10^6$, we multiply by $10^1$. To keep the number’s value the same, we must also divide the coefficient (2.0) by 10.
$$ 2.0 \times 10^5 = 0.20 \times 10^6 $$
Now our subtraction is much friendlier:
$$ (1.4 \times 10^6) – (0.20 \times 10^6) $$
Since the powers of 10 are the same, we can just subtract the coefficients:
$$ (1.4 – 0.20) \times 10^6 = 1.2 \times 10^6 $$
The result, $1.2 \times 10^6$ km, is already in proper scientific notation. Great job!
Question 5: Let’s wrap up with an equation. Can you solve for $x$? The trick is to find a common base. $$ 4^{3x-2} = 8^{x+1} $$
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Solution: This looks tough, but once you see the trick, it’s a piece of cake. The bases are different (4 and 8), so we can’t just equate the exponents yet. We need to rewrite both 4 and 8 using a common base.
What’s a number that can be raised to a power to get both 4 and 8? That’s right, it’s 2!
$4 = 2^2$
$8 = 2^3$
Now let’s substitute these back into the original equation:
$$ (2^2)^{3x-2} = (2^3)^{x+1} $$
Using the ‘power of a power’ rule (multiply the exponents), we get:
$$ 2^{2(3x-2)} = 2^{3(x+1)} $$
$$ 2^{6x-4} = 2^{3x+3} $$
Hooray! The bases are the same. This means their exponents must be equal. So we can just set up an equation with the exponents and solve for $x$.
$$ 6x – 4 = 3x + 3 $$
Let’s get the $x$ terms on one side and the numbers on the other.
$$ 6x – 3x = 3 + 4 $$
$$ 3x = 7 $$
$$ x = \frac{7}{3} $$
And that’s our final answer. You totally nailed it!
Level Up: IB Exam Style – Section A Practice
Alright, let’s turn up the heat. These questions are designed to feel like the short-answer questions you’d find in Section A of an IB exam. They might require a couple of steps to solve.
Tip
Hey! Before we jump in, let’s have a quick refresh. Remember that when you’re dealing with exponents, it’s all about the base. Your main goal is often to get the bases the same so you can apply those trusty exponent laws. For scientific notation, keep the number part separate from the power of 10 part. You’ve got this!
IB Exam Style Question
Question 1: Let’s get warmed up with some simplification. Consider the expression $\frac{(4x^8)^{1/2}}{6x^{-3}}$.
(a) Simplify $(4x^8)^{1/2}$.
(b) Hence, write the entire expression in the form $Ax^p$, where $A$ and $p$ are constants.
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Solution:
Alright, let’s break this one down step-by-step.
(a) We need to apply the power of $\frac{1}{2}$ (which is the same as taking the square root) to both parts inside the parentheses.
$(4x^8)^{1/2} = 4^{1/2} \times (x^8)^{1/2}$
The square root of 4 is 2. For the $x$ term, we multiply the powers: $8 \times \frac{1}{2} = 4$.
So, $(4x^8)^{1/2} = 2x^4$.
(b) Now we substitute our result from part (a) back into the original fraction.
$$\frac{2x^4}{6x^{-3}}$$
Let’s handle the constant coefficients first: $\frac{2}{6}$ simplifies to $\frac{1}{3}$.
Next, the variables. When we divide terms with the same base, we subtract the exponents: $x^4 \div x^{-3} = x^{4 – (-3)} = x^{4+3} = x^7$.
Putting it all together, we get $\frac{1}{3}x^7$. So, $A = \frac{1}{3}$ and $p=7$. Great job!
IB Exam Style Question
Question 2: Time to solve an equation! Find the value of $x$ in the following equation: $$25^{x-1} = \left(\frac{1}{125}\right)^{2x}$$
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Solution:
This looks tricky, but the key is to find a common base for 25 and 125. They both look like powers of 5, right?
We know that $25 = 5^2$ and $125 = 5^3$.
Let’s rewrite the term on the right side. Since $125 = 5^3$, then $\frac{1}{125} = 125^{-1} = (5^3)^{-1} = 5^{-3}$.
Now, let’s substitute these back into our original equation:
$$(5^2)^{x-1} = (5^{-3})^{2x}$$
Using the power of a power rule (multiply the exponents), we get:
$$5^{2(x-1)} = 5^{-3(2x)}$$
$$5^{2x-2} = 5^{-6x}$$
Since the bases are now the same, we can just set the exponents equal to each other!
$$2x – 2 = -6x$$
Now it’s a simple linear equation. Let’s add $6x$ to both sides:
$$8x – 2 = 0$$
Add 2 to both sides:
$$8x = 2$$
And divide by 8:
$$x = \frac{2}{8} = \frac{1}{4}$$
And that’s our answer! See? Not so bad once you find that common base.
IB Exam Style Question
Question 3: Let’s bring in some science. The mass of a single proton is approximately $1.6 \times 10^{-27}$ kg, and the mass of a single electron is $8.0 \times 10^{-31}$ kg.
(a) Find the combined mass of $3.0 \times 10^{28}$ protons.
(b) How many times more massive is a proton than an electron? Give your answer in standard form.
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Solution:
This is a perfect place to flex our scientific notation muscles. Remember to handle the numbers and the powers of 10 separately.
(a) To find the combined mass, we multiply the mass of one proton by the number of protons.
$$ \text{Mass} = (1.6 \times 10^{-27}) \times (3.0 \times 10^{28}) $$
Multiply the number parts: $1.6 \times 3.0 = 4.8$.
Multiply the power of 10 parts (add the exponents): $10^{-27} \times 10^{28} = 10^{-27+28} = 10^1$.
Combine them: $4.8 \times 10^1$ kg. This is just $48$ kg. That’s a lot of protons!
(b) To find how many times more massive a proton is, we need to divide the proton’s mass by the electron’s mass.
$$ \text{Ratio} = \frac{1.6 \times 10^{-27}}{8.0 \times 10^{-31}} $$
Divide the number parts: $1.6 \div 8.0 = 0.2$.
Divide the power of 10 parts (subtract the exponents): $10^{-27} \div 10^{-31} = 10^{-27 – (-31)} = 10^{-27+31} = 10^4$.
Combine them: $0.2 \times 10^4$.
Hold on! The question asks for the answer in standard form (which is proper scientific notation). The number part must be between 1 and 10. We need to adjust $0.2$.
$0.2 = 2 \times 10^{-1}$.
So, our answer is $(2 \times 10^{-1}) \times 10^4 = 2 \times 10^{-1+4} = 2 \times 10^3$.
So, a proton is 2000 times more massive than an electron. Cool!
IB Exam Style Question
Question 4: Don’t let this fraction scare you! Fully simplify the following expression, leaving your answer with positive exponents. $$ \frac{(2a^2b^{-1})^3}{\sqrt{16a^8b^2}} $$
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Solution:
Okay, the best strategy here is to simplify the numerator and the denominator separately first, then put them back together.
Numerator: We need to cube everything inside the parentheses.
$(2a^2b^{-1})^3 = 2^3 \times (a^2)^3 \times (b^{-1})^3 = 8 \times a^{2 \times 3} \times b^{-1 \times 3} = 8a^6b^{-3}$.
Denominator: The square root is the same as an exponent of $\frac{1}{2}$.
$\sqrt{16a^8b^2} = (16a^8b^2)^{1/2} = 16^{1/2} \times (a^8)^{1/2} \times (b^2)^{1/2} = 4 \times a^{8 \times 1/2} \times b^{2 \times 1/2} = 4a^4b^1$.
Putting it together: Now we divide our simplified numerator by our simplified denominator.
$$ \frac{8a^6b^{-3}}{4a^4b^1} $$
Handle each part: numbers, $a$’s, and $b$’s.
Numbers: $\frac{8}{4} = 2$.
$a$’s: $\frac{a^6}{a^4} = a^{6-4} = a^2$.
$b$’s: $\frac{b^{-3}}{b^1} = b^{-3-1} = b^{-4}$.
So we have $2a^2b^{-4}$. The question asks for positive exponents, so we need to move the $b^{-4}$ to the denominator.
Final Answer: $$ \frac{2a^2}{b^4} $$
You made it through! That one required careful application of a bunch of rules.
IB Exam Style Question
Question 5: Let’s think a bit more abstractly. Given that $x = 5^k$ and $y = 5^m$, express the following in the form $5^p$, where $p$ is an expression in terms of $k$ and $m$. $$ \frac{x^3}{\sqrt[3]{y}} $$
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Solution:
This is a great question to test if you really understand the exponent laws, not just how to use them with numbers. The goal is to substitute and simplify.
First, let’s rewrite the terms in the expression using the definitions of $x$ and $y$.
$x^3 = (5^k)^3$. Using the power of a power rule, this becomes $5^{3k}$.
Next, let’s deal with the cube root of $y$. A cube root is the same as an exponent of $\frac{1}{3}$.
$\sqrt[3]{y} = y^{1/3} = (5^m)^{1/3}$. Again, using the power rule, this becomes $5^{\frac{m}{3}}$.
Now, let’s substitute these back into the fraction:
$$ \frac{x^3}{\sqrt[3]{y}} = \frac{5^{3k}}{5^{m/3}} $$
We’re dividing terms with the same base (5), so we subtract the exponents.
$$ 5^{3k – \frac{m}{3}} $$
And that’s it! We’ve written the expression in the form $5^p$, where $p = 3k – \frac{m}{3}$.
Awesome work finishing this set!
The Boss Level: IB Exam Style – Section B Deep Dive
This is it—the final challenge. Section B questions are longer, multi-part problems that often combine several topics. Take your time, read carefully, and use your answers from previous parts.
IB Exam Style Question
Question: An interstellar probe, the “Odyssey,” travels from a star system to a distant nebula. The journey is split into two parts.
In the first part, the probe travels for $2.5 \times 10^8$ seconds at an average speed of $8 \times 10^7$ m/s.
- Calculate the distance covered in the first part of the journey.
- The total distance to the nebula is $5.12 \times 10^{16}$ m. Calculate the distance the probe must cover in the second part of its journey. Give your answer in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.
- The second part of the journey is covered at a much higher average speed. It takes the probe only $4 \times 10^7$ seconds to cover the remaining distance. Calculate the average speed of the probe during the second part of the journey.
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Solution:
Hey! Let’s walk through this step-by-step. This is a great example of how scientific notation makes handling huge numbers in physics problems way more manageable.
- To find the distance for the first part, we use the classic formula: distance = speed × time.
$$ \text{Distance}_1 = (8 \times 10^7 \text{ m/s}) \times (2.5 \times 10^8 \text{ s}) $$
Let’s group the numbers and the powers of 10.
$$ = (8 \times 2.5) \times (10^7 \times 10^8) $$
$$ = 20 \times 10^{7+8} = 20 \times 10^{15} \text{ m} $$
Since the question doesn’t specify scientific notation here, this is fine, but it’s good practice to write it as $2 \times 10^{16}$ m. - The remaining distance is just the total distance minus the distance from part (a).
$$ \text{Distance}_2 = \text{Total Distance} – \text{Distance}_1 $$
$$ = 5.12 \times 10^{16} – 2 \times 10^{16} $$
The powers of 10 are the same, which is awesome! That makes subtraction super easy. We just subtract the coefficients.
$$ = (5.12 – 2) \times 10^{16} = 3.12 \times 10^{16} \text{ m} $$
This is already in the correct form, so we’re all set! - For the speed in the second part, we just rearrange the formula: speed = distance / time.
$$ \text{Speed}_2 = \frac{\text{Distance}_2}{\text{Time}_2} = \frac{3.12 \times 10^{16} \text{ m}}{4 \times 10^7 \text{ s}} $$
Again, let’s separate the numbers and the powers of 10.
$$ = \left(\frac{3.12}{4}\right) \times \left(\frac{10^{16}}{10^7}\right) $$
$$ = 0.78 \times 10^{16-7} = 0.78 \times 10^9 \text{ m/s} $$
Ah, but wait! For proper scientific notation, the first number needs to be between 1 and 10. Let’s adjust it.
$$ = 7.8 \times 10^{-1} \times 10^9 = 7.8 \times 10^8 \text{ m/s} $$
And there you have it! The probe really sped up for that second leg.
IB Exam Style Question
Question: A biologist is studying a sample of extremophile bacteria. The mass of a single bacterium is approximately $1.5 \times 10^{-16}$ kg.
- The total mass of the initial sample is $3.6 \times 10^{-7}$ kg. Calculate the number of bacteria in the initial sample.
- Under special conditions, the number of bacteria in the sample triples every hour. Determine the number of bacteria in the sample after 6 hours.
- Calculate the total mass of the bacteria sample after 6 hours. Give your answer in kilograms, in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.
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Solution:
Ready to dive into some tiny biology with massive numbers? This problem is a perfect blend of scientific notation and exponential growth.
- To find the initial number of bacteria, we’ll divide the total mass of the sample by the mass of a single bacterium.
$$ \text{Number of bacteria} = \frac{\text{Total mass}}{\text{Mass of one bacterium}} = \frac{3.6 \times 10^{-7}}{1.5 \times 10^{-16}} $$
Let’s split the calculation into two parts: the coefficients and the powers of 10.
$$ = \left(\frac{3.6}{1.5}\right) \times \left(\frac{10^{-7}}{10^{-16}}\right) $$
$$ = 2.4 \times 10^{-7 – (-16)} = 2.4 \times 10^{-7+16} = 2.4 \times 10^9 $$
So, we started with 2.4 billion bacteria! - The number of bacteria triples every hour. This is exponential growth! The formula for this is $N = N_0 \times r^t$, where $N_0$ is the initial amount, $r$ is the growth rate (3, in this case), and $t$ is time in hours.
$$ \text{Number after 6 hours} = (2.4 \times 10^9) \times 3^6 $$
First, let’s figure out $3^6$. $3^2=9$, $3^4=81$, so $3^6 = 81 \times 9 = 729$.
$$ = (2.4 \times 10^9) \times 729 $$
Now we multiply $2.4 \times 729$:
$$ 2.4 \times 729 = 1749.6 $$
So the number of bacteria is $1749.6 \times 10^9$. Let’s pop that into scientific notation.
$$ 1749.6 \times 10^9 = 1.7496 \times 10^3 \times 10^9 = 1.7496 \times 10^{12} $$
That’s a lot of bacteria! - Finally, to find the new total mass, we multiply the number of bacteria after 6 hours by the mass of a single bacterium.
$$ \text{Total mass after 6 hours} = (1.7496 \times 10^{12}) \times (1.5 \times 10^{-16}) $$
Group ’em up:
$$ = (1.7496 \times 1.5) \times (10^{12} \times 10^{-16}) $$
$$ = 2.6244 \times 10^{12-16} = 2.6244 \times 10^{-4} \text{ kg} $$
This is in the correct form, so we’re done. From a tiny mass to a slightly less tiny, but still very small, mass!
IB Exam Style Question
Tip
When you’re faced with an exponential equation like in part (b), the main goal is to get the same base on both sides. Remember that roots and fractions can be written as powers! For example, $\sqrt[c]{b^a} = b^{a/c}$ and $\frac{1}{b^a} = b^{-a}$. Combining these is the key!
Question: This question tests your pure algebraic manipulation skills with exponents.
- Write the expression $\frac{(9a^8 b^{-2})^{\frac{3}{2}}}{2a^2 b}$ in the form $Ca^p b^q$, where C, p, and q are constants.
- Hence, solve the equation $\frac{(9a^8 b^{-2})^{\frac{3}{2}}}{2a^2 b} = 54$ for $a$ when $b=3$.
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Solution:
Time for some pure algebra! No stories, just exponents. Let’s flex those manipulation muscles.
- Our goal is to simplify this beast of an expression. Let’s focus on the numerator first, distributing that $\frac{3}{2}$ power to everything inside the parentheses.
$$ (9a^8 b^{-2})^{\frac{3}{2}} = 9^{\frac{3}{2}} \cdot (a^8)^{\frac{3}{2}} \cdot (b^{-2})^{\frac{3}{2}} $$
Let’s break that down piece by piece:- $9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$.
- $(a^8)^{\frac{3}{2}} = a^{8 \times \frac{3}{2}} = a^{12}$.
- $(b^{-2})^{\frac{3}{2}} = b^{-2 \times \frac{3}{2}} = b^{-3}$.
So, the numerator simplifies to $27a^{12}b^{-3}$. Now, let’s put it back into the fraction.
$$ \frac{27a^{12}b^{-3}}{2a^2 b^1} $$
Now we handle the coefficients and each variable separately. Remember to subtract the exponents of the denominator from the exponents of the numerator.
$$ = \left(\frac{27}{2}\right) a^{12-2} b^{-3-1} = \frac{27}{2}a^{10}b^{-4} $$
This is in the form $Ca^p b^q$ with $C = \frac{27}{2}$, $p=10$, and $q=-4$. Done! - The “Hence” in the question is a huge clue! It means we should use our simplified answer from part (a).
$$ \frac{27}{2}a^{10}b^{-4} = 54 $$
We are given that $b=3$. Let’s substitute that in.
$$ \frac{27}{2}a^{10}(3)^{-4} = 54 $$
Remember that $3^{-4} = \frac{1}{3^4} = \frac{1}{81}$.
$$ \frac{27}{2}a^{10} \left(\frac{1}{81}\right) = 54 $$
Let’s simplify the fraction on the left. 27 goes into 81 three times.
$$ \frac{1}{2}a^{10} \left(\frac{1}{3}\right) = 54 \implies \frac{1}{6}a^{10} = 54 $$
Now, let’s isolate $a^{10}$ by multiplying both sides by 6.
$$ a^{10} = 54 \times 6 = 324 $$
To find $a$, we take the 10th root of both sides.
$$ a = \sqrt[10]{324} $$
While you could leave it like that, we can simplify $\sqrt[10]{324}$ a bit. Notice $324=18^2$.
$$ a = (18^2)^{\frac{1}{10}} = 18^{\frac{2}{10}} = 18^{\frac{1}{5}} = \sqrt[5]{18} $$
Both $\sqrt[10]{324}$ and $\sqrt[5]{18}$ would likely be accepted. Nice work!
And that’s a wrap on this practice set! Awesome job working through these. Getting a solid handle on exponent laws and bouncing between scientific and standard notation is more than just a neat party trick—it’s a foundational skill that makes handling ridiculously huge or tiny numbers a total breeze. The more you practice manipulating these expressions, the more it becomes pure muscle memory. You’re not just solving problems; you’re building the confidence and fluency you’ll need for more complex topics down the road. Keep up the amazing work, you’re absolutely crushing it!
