$\log_{e} 5 = \ln 5 = 1.609$

6 votes

A straight line is fit to a data set (ln $x, y$). This line intercepts the abscissa at ln $x = 0.1$ and has a slope of $-0.02$. What is the value of y at $x = 5$ from the fit?

- $-0.030$
- $-0.014$
- $0.014$
- $0.030$

3 votes

Best answer

Interception at $x-$axis is $\ln \ x=0.1$ and slope $=-0.02$

Equation fo straight line is given by,

$y=mx+C$

According to the question,

$y=m(\ln \ x)+C$ where $m=-0.02$

$y=-0.02(\ln \ x)+C\quad \rightarrow(1)$

Abscissa is known as intercept in $x-$axis i.e. for $y=0,\ln \ x =0.1$

From equation $(1),$

$0=-0.02\times 0.1+C$

$C=0.002\quad \rightarrow(2)$

From equations $(1)$ and $(2),$

$y=-0.02(\ln \ x)+0.002$

At $x=5,$

$y=-0.02(\ln \ 5)+0.002$

$y=-0.030$

So, the correct answer is $(A).$

Equation fo straight line is given by,

$y=mx+C$

According to the question,

$y=m(\ln \ x)+C$ where $m=-0.02$

$y=-0.02(\ln \ x)+C\quad \rightarrow(1)$

Abscissa is known as intercept in $x-$axis i.e. for $y=0,\ln \ x =0.1$

From equation $(1),$

$0=-0.02\times 0.1+C$

$C=0.002\quad \rightarrow(2)$

From equations $(1)$ and $(2),$

$y=-0.02(\ln \ x)+0.002$

At $x=5,$

$y=-0.02(\ln \ 5)+0.002$

$y=-0.030$

So, the correct answer is $(A).$