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Which of the following logarithmic equations is equivalent to the exponential equation below?

[tex]\[ e^x = 15.29 \][/tex]

A. [tex]\(\frac{\ln 15.29}{e} = x\)[/tex]
B. [tex]\(\ln x = 15.29\)[/tex]
C. [tex]\(\ln 15.29 = x\)[/tex]
D. [tex]\(\operatorname{In} e = 15.29 x\)[/tex]


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].