Mathematics

Mathematics

Problem

For y=(x2+1)xy=(x^{2}+1)^{\sqrt{x}} with x>0x>0, which of the following expressions correctly represents ln⁡y\ln y after taking natural logarithm on both sides? a) ln⁡y=ln⁡(x2+1)x\ln y=\dfrac{\ln(x^{2}+1)}{\sqrt{x}} b) ln⁡y=(x)ln⁡(x2+1)\ln y=(\sqrt{x})^{\ln(x^{2}+1)} c) ln⁡y=x ln⁡(x2+1)\ln y=\sqrt{x}\,\ln(x^{2}+1) d) ln⁡y=x ln⁡(x2+1)\ln y=x\,\ln(x^{2}+1)
Intuitive explanation

Taking the natural logarithm of both sides transforms a power with a variable exponent into a product you can actually work with. The key insight is recognizing that ln⁡(ab)=bln⁡a\ln(a^b) = b \ln a—this lets you bring the exponent x\sqrt{x} down in front of the logarithm. Once you apply this rule to (x2+1)x(x^2+1)^{\sqrt{x}}, you'll have a clean expression for ln⁡y\ln y that's ready for differentiation or further analysis.

Intuitive explanation

Taking the natural logarithm of both sides transforms a power with a variable exponent into a product you can actually work with. The key insight is recognizing that ln⁡(ab)=bln⁡a\ln(a^b) = b \ln a—this lets you bring the exponent x\sqrt{x} down in front of the logarithm. Once you apply this rule to (x2+1)x(x^2+1)^{\sqrt{x}}, you'll have a clean expression for ln⁡y\ln y that's ready for differentiation or further analysis.

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