The function g(x)=ex2log⁡(π+x)log⁡(e+x)(x≥0) is

Since ex2 increases on [0,∞) so it is enough to consider f(x)=log⁡(π+x)log⁡(e+x)f′(x)=log⁡(e+x)×1π+x−log⁡(π+x)1e+x(log⁡(e+x))2=f′(x)=log⁡(e+x)×(e+x)−(π+x)log⁡(π+x)(π+x)(e+x)(log⁡(e+x))2Since log function is an increasing function and e<π,log⁡(e+x)0 Thus f′(x)<0 for ∀x>0⇒f(x) decreases on (0,∞)