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Question
Find the approximate value of log10 (1016), given that log10e = 0⋅4343.
Solution
Let f(x) = log10x = `(log_ex)/(log_e10)`
= (log10e)(logex)
= (0.4343) log x
∴ f'(x) = `0.4343/x`
x = 1016 = 1000 + 16 = a + h
Here, a = 1000 and h = 16
f(a) = f(1000)
= log10(1000)
= log10(10)3
= 3log10 10 ...[∵ log10 mn = n log10 m]
= 3
f'(a) = f'(1000) = `0.4343/1000` = 0.0004343
f(a + h) ≈ f(a) + hf'(a)
log10(1016) ≈ 3 + 16(0.0004343)
≈ 3 + 0.0069488
log10(1016) ≈ 3.006949
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