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What is the approximate solution to this equation?
To solve the equation [tex]\( 19 + 2 \ln x = 25 \)[/tex], follow these steps:
1. Isolate the logarithmic term: [tex]\[ 19 + 2 \ln x = 25 \][/tex] Subtract 19 from both sides: [tex]\[ 2 \ln x = 6 \][/tex]
2. Solve for [tex]\( \ln x \)[/tex]: Divide both sides by 2: [tex]\[ \ln x = 3 \][/tex]
3. Exponentiate both sides to solve for [tex]\( x \)[/tex]: Recall that [tex]\( \ln x \)[/tex] represents the natural logarithm, which is the logarithm base [tex]\( e \)[/tex]. To undo the natural logarithm, apply the exponential function (base [tex]\( e \)[/tex]) to both sides: [tex]\[ x = e^3 \][/tex]
4. Calculate the numerical value of [tex]\( e^3 \)[/tex]: Using a calculator or by approximating [tex]\( e \approx 2.718 \)[/tex]: [tex]\[ x \approx e^3 \approx 2.718^3 \approx 20.0855 \][/tex]
5. Determine which provided option is closest to the calculated value: The given options are: - A. [tex]\( x \approx 0.05 \)[/tex] - B. [tex]\( x \approx 3 \)[/tex] - C. [tex]\( x \approx 20.09 \)[/tex] - D. [tex]\( x \approx 1.93 \)[/tex]
Comparing these options with the calculated value [tex]\( x \approx 20.0855 \)[/tex]:
- 0.05 is much too small. - 3 is far too small. - 20.09 is very close to 20.0855. - 1.93 is much too small.
Therefore, the correct answer is: [tex]\[ \boxed{x \approx 20.09} \][/tex]
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