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\( \begin{array}{ll} \\ \text { Pers : } \\ 7^{3 x+5}=7^{1-x} & 2 e^{2 x}-1=5 \\ e^{2 x}=e^{3 x-1} & \ln (4 x-7)=\ln (x+11) \\ 49^{5 x+2}=\left(\frac{1}{7}\right)^{11-x} & \log _{5}(5 x+10)=4 \\ 3 e^{4 x}+9=15 & \log _{3}(x-9)+\log _{3}(x-3)=2\end{array} \)

Ask by Byrd Stephens. in the United States
Feb 25,2025

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Tutor-Verified Answer

Answer

Here are the solutions to the equations: 1. \( x = -1 \) 2. \( x = \frac{\ln 3}{2} \) 3. \( x = 1 \) 4. \( x = 6 \) 5. \( x = -\frac{5}{3} \) 6. \( x = 123 \) 7. \( x = \frac{\ln 2}{4} \) 8. \( x = 6 + 3\sqrt{2} \)

Solution

Below are one acceptable set of solutions. We solve each equation one at a time. 1) Solve 7^(3x+5) = 7^(1-x). Because the bases are the same (7), set the exponents equal:   3x + 5 = 1 - x  →  3x + x = 1 - 5  →  4x = -4  →  x = -1. 2) Solve 2e^(2x) - 1 = 5. Isolate the exponential term:   2e^(2x) = 6  →  e^(2x) = 3. Take the natural logarithm:   2x = ln 3  →  x = (ln 3)/2. 3) Solve e^(2x) = e^(3x-1). Again the bases are equal (e), so equate the exponents:   2x = 3x - 1  →  x = 1. 4) Solve ln(4x-7) = ln(x+11). Since the natural logarithms have the same base and the arguments are positive, set them equal:   4x - 7 = x + 11. Solve for x:   4x - x = 11 + 7  →  3x = 18  →  x = 6. (Verify that 4(6)-7 = 17 > 0 and 6+11 = 17 > 0.) 5) Solve 49^(5x+2) = (1/7)^(11-x). Write 49 as 7^2 and (1/7) as 7^(-1):   (7^2)^(5x+2) = 7^(2(5x+2)) = 7^(10x+4)   (1/7)^(11-x) = 7^(-(11-x)) = 7^(x-11). Now equate the exponents:   10x + 4 = x - 11  →  10x - x = -11 - 4  →  9x = -15  →  x = -15/9 = -5/3. 6) Solve log₅(5x+10) = 4. Rewrite in exponential form:   5^4 = 5x + 10. Since 5^4 = 625, we have:   5x + 10 = 625  →  5x = 615  →  x = 123. 7) Solve 3e^(4x) + 9 = 15. Isolate the exponential term:   3e^(4x) = 6  →  e^(4x) = 2. Take the natural logarithm:   4x = ln 2  →  x = (ln 2)/4. 8) Solve log₃(x-9) + log₃(x-3) = 2. Use the logarithm addition rule:   log₃[(x-9)(x-3)] = 2. Write in exponential form:   (x-9)(x-3) = 3² = 9. Expand and simplify:   x² - 12x + 27 = 9  →  x² - 12x + 18 = 0. Solve the quadratic using the quadratic formula:   x = [12 ± √(144 - 72)]/2 = [12 ± √72]/2 = [12 ± 6√2]/2 = 6 ± 3√2. Now consider the domain: the original logarithms require that x - 9 > 0 and x - 3 > 0, so x > 9. Thus, x = 6 + 3√2 (since 6 - 3√2 is less than 9). Summary of Answers:  1) x = -1  2) x = (ln 3)/2  3) x = 1  4) x = 6  5) x = -5/3  6) x = 123  7) x = (ln 2)/4  8) x = 6 + 3√2 These are the solutions to the given equations.

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Extra Insights

Get ready to tackle these equations! Solving exponential and logarithmic equations often requires a keen eye for properties of exponents and logs. For instance, remember that for any base \( a \), \( a^m = a^n \) implies \( m = n \), provided \( a \) is not equal to 1 or 0. Keeping such basics in mind can help simplify problems, especially when working with terms like \( 49^{5x+2} \) or \( e^{2x} \). A common mistake when dealing with logarithmic equations is neglecting the domain restrictions. For example, \(\ln(x + 11)\) requires \(x + 11 > 0\), meaning \(x > -11\). It's easy to overlook these limits when solving, potentially invalidating solutions. Always check for these conditions before claiming you've found a solution—it might save you from headaches later!

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