Question

If $x$ changes from $4$ to $4.01$ then find the approximate change in $log$ $x$.

Answer 1

Given,

$x=4$

On taking $log$ both sides, we get

$logx=log4$ ...... (1)

Now,

$x+\delta x=4.01$

On taking $log$ both sides, we get

$log(x+\delta x)=log4.01$ ......(2)

Divide both equations such that $\frac{equation\left(2\right)}{equation\left(1\right)}$, we get

$\frac{log(x+\delta x)}{logx}=\frac{log\left(4.01\right)}{log4}$

$log(x+\delta x-x)=log(4.01-4)$ [from the properties of $log$]

$log(x+\delta x-x)=log\left(0.01\right)$

$log\delta x=log{\left(10\right)}^{-2}$

$log\delta x=-2log10$

$=-2\times 1=-2.$

Hence, this is the answer.