Question

Find the approximate value of log 9.01 given that log 3=1.0986

Answer 1

Given,

find the approximate value of log9.01 where log3 = 1.0986

let y = logx ........(1)

differentiating with respect to x,

dy/dx = d(logx)/dx = 1/x

⇒dy/dx = 1/x

⇒dy = 1/x dx

let x = 9 , and dx = 0.01

then, dy = 1/9 × 0.01 = 0.01/9 = 0.0011 .......(2)

now log(9.01) = log(x + dx) = y + dy

⇒log(9.01) = logx + dy [ from eq (1). ]

⇒log(9.01) = log9 + 0.0011 [ Frome eq (2). ]

⇒log(9.01) = log3² + 0.0011 = 2log3 + 0.0011

⇒log(9.01) = 2 × 1.0986 + 0.0011 = 2.1983

Therefore approximate value of log(9.01) is 2.1983