Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If \(y=3^{x}\), then \(x=\sqrt[3]{y}\) b. \(\frac{\log _{b} x}{\log _{b} y}=\log _{b} x-\log _{b} y\) c. \(\log _{5} 4^{6}=4 \log _{5} 6\) d. \(2=10^{\log _{10} 2^{2}}\) e. \(2=\ln 2^{e}\) f. If \(f(x)=x^{2}+1,\) then \(f^{-1}(x)=1 /\left(x^{2}+1\right)\) g. If \(f(x)=1 / x,\) then \(f^{-1}(x)=1 / x\)

Short answer

Question: Determine whether the following statements are true or false and provide explanations or counterexamples for each. a. If \(y=3^{x}\), then \(x=\sqrt[3]{y}\) Answer: False. Counterexample: for \(y=27\), \(x=3\) but \(\sqrt[3]{27}=3\) is not equal to \(\log_3{27}\). b. \(\frac{\log _{b} x}{\log _{b} y}=\log _{b} x-\log _{b} y\) Answer: True. Both expressions are equal. c. \(\log _{5} 4^{6}=4 \log _{5} 6\) Answer: False. The correct expression is \(6\log _{5} 4=\log _{5} 4^6\). d. \(2=10^{\log _{10} 2^{2}}\) Answer: False. The correct expression is \(4 = 10^{\log_{10} 4}\). e. \(2=\ln 2^{e}\) Answer: False. The correct expression is \(2 = e\cdot\ln 2\). f. If \(f(x)=x^{2}+1,\) then \(f^{-1}(x)=1 /\left(x^{2}+1\right)\) Answer: False. The correct inverse function is \(f^{-1}(x) = \pm\sqrt{x-1}\). g. If \(f(x)=1 / x,\) then \(f^{-1}(x)=1 / x\) Answer: True. Both expressions are equal.

Step-by-step solution

Solve for x

To test the statement, express x in terms of y, based on the given equation. Do this by taking the base 3 logarithm of both sides: \(x = \log_3{y}\)

Compare the expressions

The original statement claims that \(x = \sqrt[3]{y}\). Comparing this expression to our solution of \(x = \log_3{y}\), we see that the statement is false. Provide a counterexample for further clarification: if \(y=27\), then \(x=3\), but \(\sqrt[3]{27}=3\) and not \(\log_3{27}\). b. \(\frac{\log _{b} x}{\log _{b} y}=\log _{b} x-\log _{b} y\)

Apply logarithm properties

Using the quotient rule property of logarithms, we can change the left side of the equation to a subtraction: \(\log _{b} \frac{x}{y}=\log _{b} x-\log _{b} y\)

Verify the equivalence

Comparing both sides of the equation, we find that the statement is true, as they are equal. c. \(\log _{5} 4^{6}=4 \log _{5} 6\)

Apply logarithm properties

Using the power rule property of logarithms, we can rewrite the left side of the equation: \(6\log _{5} 4=\log _{5} 4^6\)

Verify the equivalence

Comparing both sides of the equation, we find that the statement is false, as they are not equal. d. \(2=10^{\log _{10} 2^{2}}\)

Simplify the right side of the equation

First, simplify the inner exponent: \(2^2 = 4\). Then apply the logarithm to its argument: \(\log _{10} 4\). Finally, use the property that the base 10 logarithm and raising to the power of 10 are inverse operations: \(2 = 10^{\log_{10} 4}\)

Verify the equivalence

Comparing both sides of the equation, we find that the statement is false, as they are not equal. e. \(2=\ln 2^{e}\)

Apply natural logarithm properties

Using the power rule property of natural logarithms, we can rewrite the right side of the equation: \(2 = e\cdot\ln 2\)

Verify the equivalence

Comparing both sides of the equation, we find that the statement is false, as they are not equal. f. If \(f(x)=x^{2}+1,\) then \(f^{-1}(x)=1 /\left(x^{2}+1\right)\)

Find the inverse function

To find the inverse function of f(x), first replace f(x) with y. Then, switch x and y and solve for y: \(x = y^2 + 1 \Rightarrow y^2 = x-1 \Rightarrow y = \pm\sqrt{x-1}\) Therefore, \(f^{-1}(x) = \pm\sqrt{x-1}\)

Verify the equivalence

Comparing the statement's claim, we find that it is false since \(f^{-1}(x) = \pm\sqrt{x-1}\) and not \(1 / (x^2 + 1)\). g. If \(f(x)=1 / x,\) then \(f^{-1}(x)=1 / x\)

Find the inverse function

To find the inverse function of f(x), first replace f(x) with y. Then, switch x and y and solve for y: \(x = \frac{1}{y} \Rightarrow y = \frac{1}{x}\)

Verify the equivalence

Comparing the statement's claim, we find that it is true since \(f^{-1}(x) = \frac{1}{x}\).

Key concepts

Logarithm Properties

Logarithms are useful tools in mathematics, often helping to simplify complex expressions. They come with several properties that make calculations more manageable.

• **Product Rule**: The logarithm of a product is the sum of the logarithms: \(\log_b(xy) = \log_b x + \log_b y\).
• **Quotient Rule**: The logarithm of a quotient is the difference of the logarithms: \(\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y\).
• **Power Rule**: The logarithm of a power allows moving the exponent to the front: \(\log_b(x^n) = n \cdot \log_b x\).

These rules help clarify misconceptions, such as in example b. Here, applying the quotient property, \(\frac{\log_b x}{\log_b y} = \log_b x/ \log_b y\), simplifies understanding why some expressions aren't equivalent.
By using these properties, you can effectively rewrite and simplify logarithmic expressions, confirming or disproving the equivalence of given statements.

Inverse Functions

Inverse functions are used to reverse the effect of a given function. To find the inverse, you swap the variables and solve for the new dependent variable, returning the function to its original state.

• If you have \(f(x)\), its inverse \(f^{-1}(x)\) satisfies \(f(f^{-1}(x)) = x\).
• For example, if \(f(x) = x^2 + 1\), finding \(f^{-1}(x)\) involves solving \(x = y^2 + 1\), giving \(y = \pm\sqrt{x-1}\). This reveals that the given inverse statement is incorrect.

It's essential to remember that not all functions have inverses. Only bijective functions (both one-to-one and onto) possess them. Verifying an inverse is crucial for understanding the function relationship thoroughly.
With these steps, you can determine whether a proposed inverse function is valid or provide a counterexample when necessary.

Exponential Functions

Exponential functions, written as \(y = a^x\), grow rapidly and are fundamental in many scientific and financial calculations. They have properties that mirror those of logarithms, since logarithms are their inverses.

• The function \(y = a^x\) implies that for a positive base \(a\), each increase in \(x\) is equivalent to multiplying by \(a\).
• You can solve equations involving exponentials by applying logarithms to both sides, like \(x = \log_a{y}\).
• This relationship explains why the statement \(2 = 10^{\log_{10} 2^2}\) is false. The exponent doesn't match the calculated logarithm.

Exponential functions and their relations are not only beautiful in mathematics but practical in modeling real-world phenomena.
Using their properties alongside logarithmic relationships allows for the precise solution of seemingly complex problems.

Equivalence Verification

In mathematics, verifying equivalence between expressions is crucial to ensuring accuracy and understanding. When trying to prove or disprove the equivalence of two mathematical statements, you must consider all transformations and properties available.

• Equivalence can be shown by simplifying, using known mathematical properties, or substituting specific values and verifying results.
• For instance, the statement \(\log_5{4^6} eq 4 \log_5{6}\) can be disproved by simply applying the power rule correctly: \(6 \log_5{4}\), and observing the mismatch.
• Counterexamples, like showing \(y=27\) disproves \(x = \sqrt[3]{y}\) when actually should be \(x = \log_3{y}\), help to illustrate these discrepancies.

These techniques enable a clearer understanding and validation of mathematical claims, fostering a deeper insightful learning experience.
Moreover, this approach supports clarity in solving algebraic expressions, ensuring correctness and precision in mathematical solutions.