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Sagot :
To solve the exponential equation [tex]\( e^x = 15.29 \)[/tex] and find its equivalent logarithmic form, follow these steps:
1. Recognize that [tex]\( e^x \)[/tex] is an exponential function, where the base is [tex]\( e \)[/tex], the natural logarithm base. The inverse operation of exponentiation with base [tex]\( e \)[/tex] is taking the natural logarithm (ln).
2. Apply the natural logarithm to both sides of the equation to isolate the variable [tex]\( x \)[/tex]. This uses the property of logarithms that states [tex]\( \ln(e^x) = x \)[/tex] because the natural logarithm and the exponential function are inverses.
[tex]\[ \begin{align*} \ln(e^x) &= \ln(15.29) \end{align*} \][/tex]
3. Use the logarithmic identity [tex]\( \ln(e^x) = x \cdot \ln(e) \)[/tex]. Because [tex]\( \ln(e) = 1 \)[/tex], this simplifies to:
[tex]\[ \begin{align*} x \cdot \ln(e) &= \ln(15.29) \\ x \cdot 1 &= \ln(15.29) \\ x &= \ln(15.29) \end{align*} \][/tex]
Thus, the equivalent logarithmic equation for [tex]\( e^x = 15.29 \)[/tex] is:
[tex]\[ \boxed{\ln(15.29) = x} \][/tex]
Therefore, the correct option is C. [tex]\(\ln 15.29 = x\)[/tex].
1. Recognize that [tex]\( e^x \)[/tex] is an exponential function, where the base is [tex]\( e \)[/tex], the natural logarithm base. The inverse operation of exponentiation with base [tex]\( e \)[/tex] is taking the natural logarithm (ln).
2. Apply the natural logarithm to both sides of the equation to isolate the variable [tex]\( x \)[/tex]. This uses the property of logarithms that states [tex]\( \ln(e^x) = x \)[/tex] because the natural logarithm and the exponential function are inverses.
[tex]\[ \begin{align*} \ln(e^x) &= \ln(15.29) \end{align*} \][/tex]
3. Use the logarithmic identity [tex]\( \ln(e^x) = x \cdot \ln(e) \)[/tex]. Because [tex]\( \ln(e) = 1 \)[/tex], this simplifies to:
[tex]\[ \begin{align*} x \cdot \ln(e) &= \ln(15.29) \\ x \cdot 1 &= \ln(15.29) \\ x &= \ln(15.29) \end{align*} \][/tex]
Thus, the equivalent logarithmic equation for [tex]\( e^x = 15.29 \)[/tex] is:
[tex]\[ \boxed{\ln(15.29) = x} \][/tex]
Therefore, the correct option is C. [tex]\(\ln 15.29 = x\)[/tex].
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