Exponents & Logarithms - IB Mathematics Analysis and Approaches

1. Solve the equation $\log_2(x) + \log_2(x-2) = 3$.
   Answer: $x = 4$ The equation can be written as $\log_2(x(x-2)) = 3$ using the logarithm product rule. By the definition of logarithms, this means $x(x-2) = 2^3$, which simplifies to $x^2 - 2x = 8$, or $x^2 - 2x - 8 = 0$. Factoring the quadratic gives $(x-4)(x+2) = 0$, with potential solutions $x=4$ and $x=-2$. However, the domain of the logarithm function requires its argument to be positive. For $\log_2(x)$, we need $x > 0$. For $\log_2(x-2)$, we need $x-2 > 0$, which means $x > 2$. The solution $x=-2$ is extraneous as it violates these conditions. The only valid solution is $x=4$.
2. Given that $x, y > 0$ and $x, y \neq 1$, if $\log_x(y) + \log_y(x) = 2$, what is the relationship between $x$ and $y$?
   Answer: $x = y$ Let $u = \log_x(y)$. Using the change of base formula, $\log_y(x) = \frac{1}{\log_x(y)} = \frac{1}{u}$. Substituting into the given equation: $$u + \frac{1}{u} = 2$$ Multiply the entire equation by $u$ to eliminate the fraction: $$u^2 + 1 = 2u$$ Rearrange into a quadratic equation: $$u^2 - 2u + 1 = 0$$ This is a perfect square trinomial: $(u-1)^2 = 0$. The only solution is $u=1$. Substituting back $u = \log_x(y)$: $$\log_x(y) = 1$$ By the definition of logarithms, this means $y = x^1$, so $y=x$.
3. Find the sum of the solutions to the equation $9^x - 10 \cdot 3^x + 9 = 0$.
   Answer: 2 The equation can be rewritten as $(3^2)^x - 10 \cdot 3^x + 9 = 0$, which is $(3^x)^2 - 10(3^x) + 9 = 0$. Let $u = 3^x$. The equation becomes a quadratic in $u$: $u^2 - 10u + 9 = 0$. Factoring this gives $(u-1)(u-9) = 0$. So, $u=1$ or $u=9$. If $u=1$, then $3^x = 1$, which means $x=0$. If $u=9$, then $3^x = 9 = 3^2$, which means $x=2$. The solutions are $x=0$ and $x=2$. The sum of the solutions is $0 + 2 = 2$.
4. Find the value of $(\log_2 5)(\log_5 8)$.
   Answer: 3 This problem can be solved using the change of base formula for logarithms, $\log_b a = \frac{\ln a}{\ln b}$. Applying this to both terms: $$(\log_2 5)(\log_5 8) = \left(\frac{\ln 5}{\ln 2}\right) \cdot \left(\frac{\ln 8}{\ln 5}\right)$$ The $\ln 5$ terms cancel out, leaving: $$\frac{\ln 8}{\ln 2}$$ Using the change of base formula in reverse, this is equal to $\log_2 8$. Since $8 = 2^3$, $\log_2 8 = \log_2(2^3) = 3$.
5. Given that $e^x + e^{-x} = 4$, find the value of $e^{2x} + e^{-2x}$.
   Answer: 14 We are given $e^x + e^{-x} = 4$. To find $e^{2x} + e^{-2x}$, we can square the given expression. $$(e^x + e^{-x})^2 = 4^2$$ $$(e^x)^2 + 2(e^x)(e^{-x}) + (e^{-x})^2 = 16$$ Using exponent rules, $(e^x)^2 = e^{2x}$, $(e^{-x})^2 = e^{-2x}$, and $e^x \cdot e^{-x} = e^{x-x} = e^0 = 1$. So the equation becomes: $$e^{2x} + 2(1) + e^{-2x} = 16$$ $$e^{2x} + e^{-2x} + 2 = 16$$ Subtracting 2 from both sides gives: $$e^{2x} + e^{-2x} = 14$$
6. The solution to $5^{x-1} = 2 \cdot 3^x$ is $x = \log_b(c)$. Find the value of $c$.
   Answer: 10 First, we rearrange the equation to group the terms with $x$ in the exponent. $$\frac{5^x}{5} = 2 \cdot 3^x$$ $$\frac{5^x}{3^x} = 2 \cdot 5$$ $$\left(\frac{5}{3}\right)^x = 10$$ Now, we can write this exponential equation in logarithmic form: $$x = \log_{5/3}(10)$$ This is in the form $x = \log_b(c)$, where the base $b = 5/3$ and the argument $c = 10$.
7. Which of the following is equivalent to $\frac{2\ln(6) - \ln(9)}{\ln(2)}$?
   Answer: 2 First, simplify the numerator using the laws of logarithms. $$2\ln(6) - \ln(9) = \ln(6^2) - \ln(9) = \ln(36) - \ln(9)$$ Using the quotient rule, $\ln(a) - \ln(b) = \ln(a/b)$: $$\ln(36) - \ln(9) = \ln\left(\frac{36}{9}\right) = \ln(4)$$ Now the entire expression is: $$\frac{\ln(4)}{\ln(2)}$$ Using the change of base formula, $\frac{\ln(a)}{\ln(b)} = \log_b(a)$, this becomes: $$\log_2(4) = 2$$
8. Find the exact value of $\log_3 \left( \frac{1}{27\sqrt{3}} \right)$.
   Answer: $-\frac{7}{2}$ First, express the argument of the logarithm as a power of 3. We have $27 = 3^3$ and $\sqrt{3} = 3^{1/2}$. Therefore, $27\sqrt{3} = 3^3 \cdot 3^{1/2} = 3^{3 + 1/2} = 3^{7/2}$. The expression becomes $\log_3 \left( \frac{1}{3^{7/2}} \right)$. Using the rule for negative exponents, this is $\log_3(3^{-7/2})$. By the definition of logarithms, $\log_a(a^x) = x$, so the value is $-\frac{7}{2}$.
9. If $0 < a < 1$ and $b > 1$, which of the following statements must be true about $\log_a b$?
   Answer: $\log_a b < 0$ Let $y = \log_a b$. By definition, this means $a^y = b$. We are given that $0 < a < 1$ and $b > 1$. Since the base $a$ is a positive fraction less than 1, raising it to a positive power $y>0$ would result in a value less than 1 (e.g., $(1/2)^2 = 1/4$). To get a result $b$ that is greater than 1, we must raise $a$ to a negative power (e.g., $(1/2)^{-2} = 4$). Therefore, $y$ must be negative. Thus, $\log_a b < 0$. It is not necessarily less than -1, for example $\log_{1/4}(2) = -1/2$.
10. The solution to $2^{x+1} = 5^x$ is given by $x =$
    Answer: $\frac{\ln 2}{\ln 5 - \ln 2}$ To solve for $x$, we take the natural logarithm of both sides of the equation: $$\ln(2^{x+1}) = \ln(5^x)$$ Using the power rule for logarithms, $\ln(a^b) = b \ln a$: $$(x+1)\ln 2 = x\ln 5$$ Distribute $\ln 2$ on the left side: $$x\ln 2 + \ln 2 = x\ln 5$$ Rearrange the terms to isolate $x$: $$\ln 2 = x\ln 5 - x\ln 2$$ Factor out $x$ on the right side: $$\ln 2 = x(\ln 5 - \ln 2)$$ Finally, divide to solve for $x$: $$x = \frac{\ln 2}{\ln 5 - \ln 2}$$
11. Solve for $x$ in the equation $\log_4(x+2) - \log_4(x-1) = \frac{1}{2}$.
    Answer: $x = 4$ Using the quotient rule for logarithms, $\log_b(M) - \log_b(N) = \log_b(M/N)$, we can combine the terms on the left: $$\log_4\left(\frac{x+2}{x-1}\right) = \frac{1}{2}$$ By the definition of logarithms, this is equivalent to: $$\frac{x+2}{x-1} = 4^{1/2}$$ Since $4^{1/2} = \sqrt{4} = 2$: $$\frac{x+2}{x-1} = 2$$ Now, solve for $x$: $$x+2 = 2(x-1)$$ $$x+2 = 2x-2$$ $$4 = x$$ We must check that this solution is in the domain of the original logarithms. For $\log_4(x+2)$, we need $x+2 > 0$, so $x > -2$. For $\log_4(x-1)$, we need $x-1 > 0$, so $x > 1$. The solution $x=4$ satisfies both conditions.
12. If $x = \log_4(P)$ and $y = \log_{16}(P)$, which expression is equivalent to $y$?
    Answer: $\frac{x}{2}$ We can use the change of base formula. A convenient base to use is 4. $$y = \log_{16}(P) = \frac{\log_4(P)}{\log_4(16)}$$ We are given that $x = \log_4(P)$. We can calculate $\log_4(16) = \log_4(4^2) = 2$. Substituting these into the equation for $y$: $$y = \frac{x}{2}$$
13. Find the solution to the equation $2 \ln x = \ln(2x+3)$.
    Answer: $x = 3$ Using the power rule for logarithms, the left side can be rewritten: $\ln(x^2) = \ln(2x+3)$. If $\ln A = \ln B$, then $A = B$. So, we can set the arguments equal: $$x^2 = 2x+3$$ This is a quadratic equation: $x^2 - 2x - 3 = 0$. Factoring gives $(x-3)(x+1) = 0$, leading to potential solutions $x=3$ and $x=-1$. We must check these solutions in the original equation. The domain of $\ln x$ is $x > 0$. Therefore, $x=-1$ is an extraneous solution. The domain of $\ln(2x+3)$ is $2x+3>0$ or $x > -3/2$. The only solution that satisfies both domain constraints is $x=3$.
14. Given that $\ln a = 3$ and $\ln b = 4$, find the value of $\ln\left(\frac{a^2\sqrt{b}}{e}\right)$.
    Answer: 7 Using the laws of logarithms, the expression can be broken down as follows: $$\ln\left(\frac{a^2\sqrt{b}}{e}\right) = \ln(a^2) + \ln(\sqrt{b}) - \ln(e)$$ $$= 2\ln(a) + \frac{1}{2}\ln(b) - 1$$ Now, substitute the given values $\ln a = 3$ and $\ln b = 4$: $$= 2(3) + \frac{1}{2}(4) - 1$$ $$= 6 + 2 - 1 = 7$$
15. Find the value of the sum $\sum_{k=1}^{10} \log\left(\frac{k+1}{k}\right)$.
    Answer: $\log(11)$ The sum can be expanded as: $$\log\left(\frac{2}{1}\right) + \log\left(\frac{3}{2}\right) + \log\left(\frac{4}{3}\right) + \dots + \log\left(\frac{11}{10}\right)$$ Using the logarithm product rule, $\log A + \log B = \log(AB)$, we can combine these terms: $$\log\left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \dots \cdot \frac{11}{10}\right)$$ This is a telescoping product. The numerator of each term cancels with the denominator of the next term. For example, the 2 in the numerator of the first term cancels with the 2 in the denominator of the second term. This pattern continues, leaving only the denominator of the first term and the numerator of the last term: $$\log\left(\frac{11}{1}\right) = \log(11)$$
16. Find all values of $x$ that satisfy the equation $\log_x(5x-6)=2$.
    Answer: $x=2$ and $x=3$ Convert the logarithmic equation to its exponential form: $x^2 = 5x-6$. Rearrange this into a quadratic equation: $x^2 - 5x + 6 = 0$. Factor the quadratic: $(x-2)(x-3) = 0$. The potential solutions are $x=2$ and $x=3$. We must check these solutions against the conditions for logarithms. The base of a logarithm, $x$, must be positive and not equal to 1 ($x>0, x \neq 1$). Both $x=2$ and $x=3$ satisfy this. The argument of the logarithm, $5x-6$, must be positive. For $x=2$: $5(2)-6 = 10-6 = 4 > 0$. So, $x=2$ is a valid solution. For $x=3$: $5(3)-6 = 15-6 = 9 > 0$. So, $x=3$ is a valid solution. Both values satisfy the equation and the domain constraints.
17. If $2^a = 5$ and $5^b = 8$, find the value of $ab$.
    Answer: 3 We can solve this using substitution. We are given $2^a = 5$. Substitute this expression for 5 into the second equation: $$(2^a)^b = 8$$ Using the power rule for exponents, $(x^m)^n = x^{mn}$: $$2^{ab} = 8$$ To solve for $ab$, we express 8 as a power of 2: $$2^{ab} = 2^3$$ Since the bases are equal, the exponents must be equal. Therefore, $ab = 3$.
18. Express $\log_9(12)$ in terms of $a$, where $a = \log_3(2)$.
    Answer: $\frac{a+1}{2}$ We use the change of base formula to convert $\log_9(12)$ to base 3. $$\log_9(12) = \frac{\log_3(12)}{\log_3(9)}$$ The denominator is $\log_3(9) = \log_3(3^2) = 2$. For the numerator, we use the product rule: $\log_3(12) = \log_3(4 \cdot 3) = \log_3(4) + \log_3(3)$. This gives us $\log_3(2^2) + 1 = 2\log_3(2) + 1 = 2a + 1$. Therefore: $\log_9(12) = \frac{2a+1}{2} = \frac{a+1}{2}$.
19. The solution to $3^{2x-1} = 4$ is $x = \log_b(c)$. Find $b$ and $c$.
    Answer: $b=9, c=12$ We start by isolating the term with the exponent $x$. $$3^{2x-1} = 4$$ $$\frac{3^{2x}}{3^1} = 4$$ $$3^{2x} = 12$$ Using the exponent rule $(a^m)^n = a^{mn}$, we can rewrite the left side: $$(3^2)^x = 12$$ $$9^x = 12$$ Now, we convert this exponential equation to logarithmic form $y = \log_b(a) \iff b^y=a$. $$x = \log_9(12)$$ Comparing this to the form $x = \log_b(c)$, we have $b=9$ and $c=12$.
20. Find the value of $e^{3 \ln 2} - \log(100)$.
    Answer: 6 We evaluate each term separately. For the first term, use the logarithm power rule $n \ln a = \ln(a^n)$ and the identity $e^{\ln x} = x$. $$e^{3 \ln 2} = e^{\ln(2^3)} = e^{\ln 8} = 8$$ For the second term, $\log(100)$ means $\log_{10}(100)$. $$\log_{10}(100) = \log_{10}(10^2) = 2$$ Now, subtract the second value from the first: $$8 - 2 = 6$$

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