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