Consider the exponential function with positive base \(a\), where \(a > 0\) and \(a
eq 1\). In this question, you will explore the extended properties of exponents.
(a) Prove that for any real numbers \(x\) and \(y\), the identity \(a^x \cdot a^y = a^{x+y}\) holds by taking the natural logarithm of both sides and using the properties of logarithms. In your proof, also explain briefly how the case \(x = -y\) leads to the reciprocal property of exponentials.
(b) Using the property proven in part (a), solve the equation \(5^{2x} \cdot 5^{-x} = 125\) for \(x\).
(c) Reflect on how the proven exponential property facilitates the process of solving exponential equations and transforming multiplicative relationships into additive ones in logarithmic form.
Question
Prove that for any positive number \(a\) (with \(a\neq 1\)) and any real numbers \(x\) and \(y\), the identity \(a^x \cdot a^y = a^{x+y}\) holds by taking the natural logarithm of both sides. Additionally, show how setting \(x = -y\) leads to the reciprocal property of exponentials.
Hint
Hint
Hint
Remember to use the property \(\ln(a^k) = k\ln(a)\) and then equate the results.
-
Answer
Answer
-
Mark Scheme
Mark Scheme
-
Explanation
Explanation
- Formula Booklet
Complete Answer
By taking the natural logarithm of both sides and applying logarithm properties, we can show that if \(\ln(a^x \cdot a^y) = \ln(a^{x+y})\), then \(a^x \cdot a^y = a^{x+y}\). Setting \(y = -x\) in \(a^x \cdot a^y = a^{x+y}\) gives \(a^x \cdot a^{-x} = a^{x-x} = a^0 = 1\), hence \(a^{-x} = \frac{1}{a^x}\).
-
Mark Scheme
Mark Scheme
-
Answer Explanation
Answer Explanation
Question
Solve the equation \(5^{2x} \cdot 5^{-x} = 125\) using the property proven in part (a).
Hint
Hint
Hint
Remember that \(5^{2x} \cdot 5^{-x}=5^{2x-x}=5^x\) and express 125 as a power of 5.
-
Answer
Answer
-
Mark Scheme
Mark Scheme
-
Explanation
Explanation
- Formula Booklet
-
Mark Scheme
Mark Scheme
-
Answer Explanation
Answer Explanation
Question
Explain how the exponential property \(a^x \cdot a^y = a^{x+y}\) assists in simplifying and solving exponential equations, particularly in converting multiplicative relationships into additive ones in logarithmic form.
Hint
Hint
Hint
Reflect on the inverse relationship between exponentials and logarithms.
-
Answer
Answer
-
Mark Scheme
Mark Scheme
-
Explanation
Explanation
- Formula Booklet
Complete Answer
This property is fundamental because it allows exponential equations to be simplified by converting multiplicative operations into addition through logarithms. This conversion is key in solving equations and in many algebraic manipulations.
-
Mark Scheme
Mark Scheme
-
Answer Explanation
Answer Explanation
