Answer :

log1010.1, it being given that log10e = 0.4343


Let us assume that f(x) = log10x


Also, let x = 10 so that x + Δx = 10.1


10 + Δx = 10.1


Δx = 0.1


On differentiating f(x) with respect to x, we get







We know




When x = 10, we have



Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as



Here, and Δx = 0.1


Δf = (0.04343)(0.1)


Δf = 0.004343


Now, we have f(10.1) = f(10) + Δf


f(10.1) = log1010 + 0.004343


f(10.1) = 1 + 0.004343 [ logaa = 1]


f(10.1) = 1.004343


Thus, log1010.1 ≈ 1.004343


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