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What is the approximate solution to this equation?

[tex]e^{5x} = 20[/tex]

A. [tex]x \approx 0.28[/tex]
B. [tex]x \approx 0.22[/tex]
C. [tex]x \approx 1.86[/tex]
D. [tex]x \approx 0.40[/tex]


Sagot :

To solve the equation [tex]\( e^{5x} = 20 \)[/tex]:

1. Take the natural logarithm (ln) of both sides. The natural logarithm is the inverse operation of the exponential function, which allows us to deal with the exponent more easily.
[tex]\[ \ln(e^{5x}) = \ln(20) \][/tex]

2. Apply the logarithm rule which states [tex]\( \ln(e^y) = y \)[/tex]. This simplifies our equation:
[tex]\[ 5x = \ln(20) \][/tex]

3. Isolate [tex]\( x \)[/tex] by dividing both sides of the equation by 5:
[tex]\[ x = \frac{\ln(20)}{5} \][/tex]

4. Calculate the value.
The natural logarithm of 20 (approximately [tex]\( \ln(20) \approx 2.9957 \)[/tex]):
[tex]\[ x \approx \frac{2.9957}{5} \approx 0.6 \][/tex]

Therefore, the approximate solution to the equation [tex]\( e^{5x} = 20 \)[/tex] is:

Option A: [tex]\( x \approx 0.6 \)[/tex]