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CBSE Class 12-science Answered

Using differentials find the approx value Log e 10.02 given that loge 10 = 2.3026
Asked by haroonrashidgkp | 08 Sep, 2018, 04:22: PM
Expert Answer
f(x) = loge x
a = 10 and h = 0.02
f(a) = f(10) = loge 10 = 2.3026
f'(x) = 1/x 
f'(10) = 1/10
Using approximation
f(10.02) ≈ f(10) + 0.02 f'(10)
             ≈ 2.3026 + 0.02 × 1/10 = 2.3026 + 0.002 = 2.3046
Answered by Sneha shidid | 09 Sep, 2018, 06:28: PM
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