Question

If $f(x,y)=\left(ln{x}^2\right)\left(e^{2y}\right)$, what is the approximate value of $f(2,\sqrt{2})$?

• A. 23.45
• B. 24.35
• C. 25.34
• D. 25.43
• E. 27.25

Answer 1

Answer: (A)

23.45

• Given $f\left(x\right)=\left(ln{x}^2\right)\left(e^{2y}\right)$
• So the value of $f(2,\sqrt{2})$ is $\left(ln4\right)\left(e^{2\sqrt{2}}\right)=23.45$