𝒄(𝒙) = 𝐥𝐧 𝒙
𝟑 − 𝟒
Ans:-
c(x)=lnx3−4c(x)=lnx^3-4c(x)=lnx3−4
⇒c(x)=3×lnx−4\Rightarrow c(x)= 3\times lnx -4⇒c(x)=3×lnx−4
We know that lnx=(x−1)−(x−1)22+(x−1)33−(x−1)44+........lnx=(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}-\dfrac{(x-1)^4}{4}+........lnx=(x−1)−2(x−1)2+3(x−1)3−4(x−1)4+........
So
⇒\Rightarrow⇒ c(x)=3×[(x−1)−(x−1)22+(x−1)33−(x−1)44+........]−4c(x)=3\times[(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}-\dfrac{(x-1)^4}{4}+........]-4c(x)=3×[(x−1)−2(x−1)2+3(x−1)3−4(x−1)4+........]−4
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