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प्रश्न
Assuming log10 e = 0.4343, find an approximate value of Iog10 1003
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उत्तर
Let f(x) = log 10 x then
f'(x) = `1/x` log10 e(log10 x = log10 e loge x)
f(x + Δx) – f(x) = f ‘(x) Δ
f(1003) – f(1000) = `0.4344/1000 xx 3`
log10 1003 – log10 1000 = 0.0013029
log10 1003 = log10 103 + 0.0013029
= 3 + 0.0013029
= 3.0013029
Approximate value of log10 1003 = 3.0013029
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