Exponential & Logarithmic Functions - IB Mathematics Analysis and Approaches

1. Find the exact solution to the equation $3^{2x} = 5^{x-1}$.
   Answer: $\frac{\ln 5}{\ln 5 - 2\ln 3}$ Take the natural logarithm of both sides of the equation: $\ln(3^{2x}) = \ln(5^{x-1})$. Using the power rule for logarithms, this becomes: $2x \ln 3 = (x-1) \ln 5$. Distribute $\ln 5$ on the right side: $2x \ln 3 = x \ln 5 - \ln 5$. Rearrange the equation to group terms with $x$: $\ln 5 = x \ln 5 - 2x \ln 3$. Factor out $x$: $\ln 5 = x(\ln 5 - 2\ln 3)$. Finally, solve for $x$: $x = \frac{\ln 5}{\ln 5 - 2\ln 3}$.
2. Find the sum of the solutions to the equation $(\log_3 x)^2 - \log_3(x^2) = 3$.
   Answer: $\frac{82}{3}$ The equation is $(\log_3 x)^2 - 2\log_3 x = 3$. Let $u = \log_3 x$. The equation becomes a quadratic in $u$: $u^2 - 2u - 3 = 0$. Factoring gives $(u-3)(u+1) = 0$, so $u=3$ or $u=-1$. If $u=3$, then $\log_3 x = 3 \Rightarrow x = 3^3 = 27$. If $u=-1$, then $\log_3 x = -1 \Rightarrow x = 3^{-1} = \frac{1}{3}$. Both solutions are valid as they are positive. The sum of the solutions is $27 + \frac{1}{3} = \frac{81}{3} + \frac{1}{3} = \frac{82}{3}$.
3. If $e^{2x} = 9$ and $e^{3y} = 8$, what is the value of $e^{x-y}$?
   Answer: $\frac{3}{2}$ First, solve for $e^x$ and $e^y$. From $e^{2x} = 9$, we have $(e^x)^2 = 3^2$. Since $e^x$ must be positive, $e^x = 3$. From $e^{3y} = 8$, we have $(e^y)^3 = 2^3$. Taking the cube root, we get $e^y = 2$. The expression $e^{x-y}$ can be written as $\frac{e^x}{e^y}$. Substituting the values we found: $e^{x-y} = \frac{3}{2}$.
4. A population of bacteria is modelled by $P(t) = P_0 e^{0.15t}$, where $t$ is the time in hours. Determine the time it takes for the population to triple, giving the answer in terms of natural logarithms.
   Answer: $\frac{20 \ln 3}{3}$ We need to find the time $t$ when the population $P(t)$ is three times the initial population $P_0$. So, we set $P(t) = 3P_0$. $3P_0 = P_0 e^{0.15t}$. Divide by $P_0$: $3 = e^{0.15t}$. Take the natural logarithm of both sides: $\ln 3 = 0.15t$. Solve for $t$: $t = \frac{\ln 3}{0.15}$. To express this in the form given in the options, convert the decimal to a fraction: $0.15 = \frac{15}{100} = \frac{3}{20}$. So, $t = \frac{\ln 3}{3/20} = \frac{20 \ln 3}{3}$.
5. Find the solution to the equation $\log_6(x+5) + \log_6(x) = 2$.
   Answer: 4 First, combine the logarithms using the product rule: $\log_6(x(x+5)) = 2$. The domain requires $x>0$ and $x+5>0$, so $x>0$. Convert to exponential form: $x(x+5) = 6^2 = 36$. This gives $x^2 + 5x - 36 = 0$. Factoring: $(x+9)(x-4) = 0$. The potential solutions are $x = -9$ and $x = 4$. Only $x = 4$ satisfies the domain condition $x > 0$.
6. If $\log_a b = 3$ and $\log_b c = -2$, what is the value of $\log_c a$?
   Answer: $-\frac{1}{6}$ We can use the change of base property $\log_y x = \frac{1}{\log_x y}$ and the chain rule $\log_z x = \log_z y \cdot \log_y x$. We want to find $\log_c a$. We can write $\log_c a = \log_c b \cdot \log_b a$. From the given information, $\log_a b = 3 \implies \log_b a = \frac{1}{3}$. And $\log_b c = -2 \implies \log_c b = \frac{1}{-2} = -\frac{1}{2}$. Substituting these into the expression for $\log_c a$ gives: $\log_c a = (-\frac{1}{2}) \cdot (\frac{1}{3}) = -\frac{1}{6}$.
7. Given that $\log_3(a) = p$ and $\log_3(b) = q$, express $\log_3(9a^2/b)$ in terms of $p$ and $q$.
   Answer: $2+2p-q$ We use the laws of logarithms to expand the expression. Using the quotient rule: $\log_3(9a^2/b) = \log_3(9a^2) - \log_3(b)$. Using the product rule on the first term: $\log_3(9) + \log_3(a^2) - \log_3(b)$. Using the power rule on the second term: $\log_3(9) + 2\log_3(a) - \log_3(b)$. We know that $\log_3(9) = 2$, and we are given $\log_3(a) = p$ and $\log_3(b) = q$. Substituting these values gives: $2 + 2p - q$.
8. Solve the equation $\log_2(x+1) - \log_2(x-1) = 3$.
   Answer: $\frac{9}{7}$ The domain requires $x+1>0$ and $x-1>0$, which simplifies to $x>1$. Using the quotient rule for logarithms, the equation becomes $\log_2\left(\frac{x+1}{x-1}\right) = 3$. Converting to exponential form: $\frac{x+1}{x-1} = 2^3 = 8$. Now, solve for $x$: $x+1 = 8(x-1)$ $x+1 = 8x - 8$ $9 = 7x$ $x = \frac{9}{7}$. We check if this solution is in the domain. Since $\frac{9}{7} \approx 1.28$, it is greater than 1, so the solution is valid.
9. The intensity $I$ of a light beam after passing through a substance of thickness $x$ cm is given by $I(x) = I_0 (10^{-kx})$, where $I_0$ is the initial intensity and $k$ is a constant. If the intensity is halved after passing through 2 cm of the substance, what is the value of $k$?
   Answer: $\frac{1}{2}\log 2$ When you encounter exponential decay problems involving logarithms, you're dealing with situations where a quantity decreases by a constant percentage over equal intervals. The key is translating the given condition into an equation you can solve. Given that the intensity is halved after passing through 2 cm, you know that $I(2) = \frac{I_0}{2}$. Substituting into the formula: $\frac{I_0}{2} = I_0(10^{-2k})$ Dividing both sides by $I_0$: $\frac{1}{2} = 10^{-2k}$ To solve for $k$, take the logarithm base 10 of both sides: $\log\left(\frac{1}{2}\right) = \log(10^{-2k}) = -2k$ Since $\log\left(\frac{1}{2}\right) = -\log 2$, you get: $-\log 2 = -2k$ Therefore: $k = \frac{\log 2}{2} = \frac{1}{2}\log 2$ Looking at the wrong answers: Choice (A) $\frac{1}{2}$ ignores the logarithmic relationship entirely. Choice (B) $\log 2$ comes from forgetting to divide by 2 when solving $-\log 2 = -2k$. Choice (C) $2\log 2$ results from incorrectly multiplying instead of dividing by 2. The correct answer is (D) $\frac{1}{2}\log 2$. Study tip: In exponential decay problems, always set up your equation using the given condition first, then use logarithms to solve for unknown constants. Remember that $\log(a^n) = n\log(a)$ is crucial for extracting variables from exponents.
10. Given that $2^{2x+1} - 5(2^x) + 2 = 0$, find the sum of all possible values of $x$.
    Answer: 0 The equation can be rewritten as $2 \cdot 2^{2x} - 5(2^x) + 2 = 0$, which is $2(2^x)^2 - 5(2^x) + 2 = 0$. Let $u = 2^x$. The equation becomes $2u^2 - 5u + 2 = 0$. Factoring this quadratic gives $(2u-1)(u-2) = 0$. So, $u = \frac{1}{2}$ or $u=2$. If $u = \frac{1}{2}$, then $2^x = \frac{1}{2} = 2^{-1}$, so $x=-1$. If $u=2$, then $2^x = 2^1$, so $x=1$. The sum of the solutions is $-1 + 1 = 0$.
11. The function $f(x) = e^{2x} - 6e^x + 5$ has $x$-intercepts at $A$ and $B$. Find the exact distance between $A$ and $B$.
    Answer: $\ln 5$ When you encounter an exponential function with $e^x$ terms, look for opportunities to use substitution to transform it into a more familiar form. This function has the structure of a quadratic in disguise. To find the x-intercepts, set $f(x) = 0$: $e^{2x} - 6e^x + 5 = 0$. Notice that $e^{2x} = (e^x)^2$, so let $u = e^x$. This transforms our equation into the quadratic: $u^2 - 6u + 5 = 0$. Factoring: $(u - 1)(u - 5) = 0$, so $u = 1$ or $u = 5$. Substituting back: $e^x = 1$ or $e^x = 5$. Since $e^x = 1$ means $x = 0$, and $e^x = 5$ means $x = \ln 5$, our x-intercepts are at $x = 0$ and $x = \ln 5$. The distance between these points is $|\ln 5 - 0| = \ln 5$. Choice A (4) might tempt you if you incorrectly solved $e^x = 5$ as $x = 4$, forgetting that you need the natural logarithm. Choice B ($\ln 4$) could result from arithmetic errors in the quadratic, perhaps getting $u = 4$ instead of $u = 5$. Choice C (5) represents the trap of confusing the solution $e^x = 5$ with $x = 5$, rather than taking the logarithm. Strategy tip: When you see exponential equations with multiple $e^x$ terms, always check if substitution $u = e^x$ creates a quadratic. Remember that solving $e^x = k$ requires taking the natural logarithm: $x = \ln k$.
12. If $\log_x 81 = 4$ and $\log_3 y = -1$, find the value of $xy$.
    Answer: 1 We need to solve for $x$ and $y$ separately. From $\log_x 81 = 4$, we can write this in exponential form as $x^4 = 81$. Since $81 = 3^4$, we have $x^4 = 3^4$. As the base $x$ of a logarithm must be positive (and not equal to 1), we have $x=3$. From $\log_3 y = -1$, we can write this in exponential form as $y = 3^{-1} = \frac{1}{3}$. Now, we find the product $xy$: $xy = 3 \cdot \frac{1}{3} = 1$.
13. Find the set of values of $x$ for which $2 \log_5(x-1) > \log_5(x+5)$.
    Answer: $\{x \in \mathbb{R} \mid x > 4\}$ First, establish the domain for the logarithms. We need $x-1 > 0 \Rightarrow x > 1$ and $x+5 > 0 \Rightarrow x > -5$. The combined domain is $x > 1$. Now, solve the inequality: $\log_5((x-1)^2) > \log_5(x+5)$. Since $\log_5 x$ is an increasing function, we can remove the logarithms: $(x-1)^2 > x+5$. $x^2 - 2x + 1 > x+5$ $x^2 - 3x - 4 > 0$ $(x-4)(x+1) > 0$. The roots of the quadratic are $x=4$ and $x=-1$. The inequality holds when $x < -1$ or $x > 4$. Finally, we must find the intersection of this solution set with the domain $x > 1$. The only values that satisfy both conditions are $x > 4$.
14. Find the value of $x$ that satisfies the equation $\log_4 x + \log_{16} x = 3$.
    Answer: 16 Using the change of base formula, we can write $\log_{16} x = \frac{\log_4 x}{\log_4 16}$. Since $\log_4 16 = 2$, the equation becomes $\log_4 x + \frac{\log_4 x}{2} = 3$. Combining the terms on the left gives $\frac{3}{2} \log_4 x = 3$. Multiplying by $\frac{2}{3}$ gives $\log_4 x = 2$. Therefore, $x = 4^2 = 16$.
15. The numbers $\ln(x-2)$, $\ln(x)$, and $\ln(x+6)$ are three consecutive terms of an arithmetic sequence. Find the value of $x$.
    Answer: 3 For an arithmetic sequence, the common difference between consecutive terms is constant. So, $\ln(x) - \ln(x-2) = \ln(x+6) - \ln(x)$. This can be rewritten as $2\ln(x) = \ln(x-2) + \ln(x+6)$. The domain requires $x-2>0$, $x>0$, and $x+6>0$, which simplifies to $x>2$. Using logarithm laws, the equation becomes $\ln(x^2) = \ln((x-2)(x+6))$. Equating the arguments gives $x^2 = (x-2)(x+6) = x^2 + 4x - 12$. Subtracting $x^2$ from both sides gives $0 = 4x - 12$, which means $4x = 12$ and $x=3$. This solution is valid as it satisfies the domain condition $x>2$.
16. Find the sum of the solutions of the equation $4^x - 6 \cdot 2^x + 8 = 0$.
    Answer: 3 Rewrite the equation using the base 2. Since $4^x = (2^2)^x = (2^x)^2$, the equation is $(2^x)^2 - 6 \cdot 2^x + 8 = 0$. Let $u = 2^x$. The equation becomes a quadratic: $u^2 - 6u + 8 = 0$. Factoring gives $(u-2)(u-4) = 0$, so the solutions for $u$ are $u=2$ and $u=4$. Let the solutions for $x$ be $x_1$ and $x_2$. If $u=2$, then $2^{x_1} = 2 \implies x_1 = 1$. If $u=4$, then $2^{x_2} = 4 = 2^2 \implies x_2 = 2$. The sum of the solutions is $x_1 + x_2 = 1 + 2 = 3$.
17. Given that $y = \log_a x$, which expression is equivalent to $\log_a (ax)^3$?
    Answer: $3+3y$ We use the properties of logarithms. First, the power rule: $\log_a (ax)^3 = 3 \log_a (ax)$. Next, the product rule: $3 \log_a (ax) = 3 (\log_a a + \log_a x)$. We know that $\log_a a = 1$ and we are given that $y = \log_a x$. Substituting these in gives: $3 (1 + y) = 3 + 3y$.
18. Find the value of the car after 5 years, to the nearest dollar.
    Answer: $11664 First, find the decay constant $k$. We have $V(2) = 18000$, so $18000 = 25000 e^{-2k}$. This gives $\frac{18}{25} = e^{-2k}$, so $e^{-k} = \sqrt{\frac{18}{25}} = \frac{3\sqrt{2}}{5}$. Now find $V(5) = 25000 e^{-5k} = 25000(e^{-k})^5 = 25000\left(\frac{3\sqrt{2}}{5}\right)^5 = 25000 \cdot \frac{243 \cdot 4\sqrt{2}}{3125} = 25000 \cdot \frac{972\sqrt{2}}{3125} \approx 11664$. To the nearest dollar, the value is $11664.
19. The equation $9^x - 5 \cdot 3^{x+1} + 54 = 0$ has two solutions, $x_1$ and $x_2$. Find the value of $x_1 + x_2$.
    Answer: $\log_3 54$ First, rewrite the equation in terms of $3^x$. $9^x = (3^2)^x = (3^x)^2$ and $3^{x+1} = 3 \cdot 3^x$. The equation becomes $(3^x)^2 - 5(3 \cdot 3^x) + 54 = 0$, which simplifies to $(3^x)^2 - 15(3^x) + 54 = 0$. Let $u = 3^x$. The equation is $u^2 - 15u + 54 = 0$. Factoring the quadratic gives $(u-6)(u-9)=0$. The solutions for $u$ are $u=6$ and $u=9$. So, $3^{x_1} = 6$ and $3^{x_2} = 9$. This gives $x_1 = \log_3 6$ and $x_2 = \log_3 9 = 2$. The sum of the solutions is $x_1 + x_2 = \log_3 6 + \log_3 9$. Using the product rule for logarithms, this is $\log_3(6 \cdot 9) = \log_3 54$.
20. Find the solution to the equation $e^x - 12e^{-x} = 1$.
    Answer: $\ln 4$ Let $u = e^x$. Since $e^{-x} = 1/e^x = 1/u$, the equation becomes $u - \frac{12}{u} = 1$. To eliminate the fraction, multiply the entire equation by $u$ (note that $u=e^x$ is always positive): $u^2 - 12 = u$. Rearrange into a standard quadratic form: $u^2 - u - 12 = 0$. Factor the quadratic: $(u-4)(u+3) = 0$. The solutions for $u$ are $u=4$ and $u=-3$. Since $u = e^x$, $u$ must be positive. Therefore, $u=-3$ is not a valid solution. We are left with $e^x = 4$. Solving for $x$ gives $x = \ln 4$.

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