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\( \begin{array}{llll}\text { (1) } 5.9^{x-1}=5 & \text { (2) } 7.49^{x-2}=49 & \text { (3) } & \frac{1}{49^{x}}=343 \\ \text { (4) } 5.125^{x+3}=\frac{1}{25} & \text { (5) }\left(\frac{2}{3}\right)^{x-2}=\frac{8}{27} & \text { (6) } & (0,25)^{x}=0,125 \\ \text { (7) }(0,2)^{x-2}=0,04 & \text { (8) } 0,4^{x}=0,064 & \text { (9) } & \left(3^{x+1}\right)^{3}=9^{x-3} \\ \text { (10) } 4^{x}=16^{x-1} & \text { (11) } 3^{x} \cdot 9^{x-1}=81 & \text { (12) } & (0,5)^{x-1}=4^{-x} \\ \text { (13) } 8 i^{2 x+1}=27^{x-2} & \text { (14) } 8^{-x}=2.4^{x-1} & \text { (15) } & 4.2^{x}=(0,5)^{x-2}\end{array} \) (c) Solve the following equations:

Ask by Mcfarlane Macdonald. in South Africa
Mar 09,2025

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Here are the solutions to the equations: 1. \( x = \log_{5.9}(5) + 1 \) 2. \( x = 2\log_{7.49}(7) + 2 \) 3. \( x = -1.5 \) or \( x = -1.5 \) 4. \( x = -3\log_{5.125}(5) - 3 \) 5. \( x = 5 \) 6. \( x = 1.5 \) or \( x = 1.5 \) 7. \( x = 4 \) 8. \( x = 3 \) 9. \( x = -9 \) 10. \( x = 2 \) 11. \( x = 2 \) 12. \( x = -1 \) 13. Cannot solve with the given information. 14. \( x = \log_{2.4}(12) - \log_{2.4}(5) \) 15. \( x = 2\log_{4.2}(2) \)

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Bonus Knowledge

To solve these equations, it's crucial to recognize patterns in exponential rules. For instance, remember \(a^{m} = a^{n} \implies m = n\) if the bases are the same. For problems where bases differ, consider rewriting them in terms of powers of a base, such as expressing each term as powers of 2 or 3 when relevant. Another engaging trick is to convert equations into logarithmic form, which can simplify your solutions! For example, in an equation involving an exponential, \(a^b = c\) can be rewritten as \(b = \log_a(c)\). Common mistakes involve failing to check whether the base is greater than zero or not equal to one, as that can invalidate the rules you apply.

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