IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Number-Laws of exponents
Topic :Number- Weightage : 21 %
All Questions for Topic : Laws of exponents, including fractional/rational exponents Logarithms, including laws of logarithms and the use of technology to find values Upper and lower bounds
Question (a)
▶️Answer/Explanation
Ans:
1. \( \sqrt{x^3} \) is equivalent to \( x^{\frac{3}{2}} \).
2. \( \sqrt[3]{x^2} \) is equivalent to \( x^{\frac{2}{3}} \).
3. \( \frac{x^2}{3} \) remains not equivalent to any of the provided expressions.
Question (b)
▶️Answer/Explanation
Ans:
Expression: \(\frac{9^x \times 3^{2x}}{3^x}\)
Here, we have a combination of exponential terms and division involving the same base, which is 3. To simplify this expression, we can use the properties of exponents:
1. \(9^x\) can be written as \(3^{2x}\), since \(9 = 3^2\).
2. Now we have \(3^{2x} \times 3^{2x}\) in the numerator and \(3^x\) in the denominator. We can combine the bases by adding the exponents.
\(3^{2x} \times 3^{2x} = 3^{2x + 2x} = 3^{4x}\).
3. So, the expression becomes \(\frac{3^{4x}}{3^x}\).
4. When you have the same base and you’re dividing, you can subtract the exponents: \(3^{4x – x} = 3^{3x}\).
Now, based on the simplified expression \(3^{3x}\), let’s choose the correct option from the draggable items:
- \(9^{2x}\)
- \(9^x\)
- \(27^x\)
The correct option is: \(27^x\)