Question

If log103 = 0.4771 and log10e = 0.4343, then what is the value of log1030.5?

(a) 1.43

(b) 1.5

(c) 1.484

(d) 1.4

The question was asked in an interview.

My question comes from Application of Derivative for Error Determination in section Application of Derivatives of Mathematics – Class 12

Answer 1

Right option is (c) 1.484

The explanation: Let, y = log10x

Then f’(x) = d/dx[log10x]

= d/dx[logex * log10e]

= 1/x(log10e)

Now, f(x + δx) = f(x) + f’(x) δx

Putting x = 30 and δx = 0.5 in the above equation we get,

f(30 + 0.5) = f(30) + f’(30) δx

=> f(30.5) = log1030 + (1/30) log10e * 0.5

Putting the values we get,

log1030.5 = 1.4843 = 1.484