Q8 of 292 Page 469

find the approximate values of 10.02, given that 10 = 2.3026

Let


Let x =10 and Δx = 0.02.


As


⇒


We, know


⇒


∴


⇒


⇒


∴ Δy = 0.002


Also,


Δy = f(x+Δx)-f(x)


∴ 0.002 = loge(10+0.02) – loge(10)


⇒ 0.002 = loge(10.02) – 2.3026


⇒ loge(10.02) = 2.3046.


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