Question

using Differential, Find the Approximate Value of the Log base 10 (10.1), It Being Given that Log base 10(e) = 0.4343 ?

Answer 1

Step-by-step explanation:

Let y = log10 X

y = log e x/ log e10

dy/dx = 0.4343/x

delta y = dy/dx. delta x

= 0.4343/x.0.04

put x= 4

0.4343× 0.04/4

= 0.004343

so y + delta y = log 10 (4.04) = 0.6021+ 0.004343= 0.606443

abcd = 6064

Answer 2

Answer:

Refer to this picture

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