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**Hint:**Plot the graph of the given $\ln x$ and $\ln y$ using given coordinates.

Find the slope of the straight line formed by using a two-point formula.

In the Cartesian plane, the horizontal line is the x-axis, and the vertical line is the y-axis. The x-axis and y-axis are perpendicular to each other.

The point where the x-axis and y-axis cut each other is called the origin, $\left( {0,0} \right)$ .

The coordinates of a point of the Cartesian plane is written in the form $\left( {x,y} \right)$.

To plot coordinates $\left( {1,2} \right)$ on the Cartesian plane, from the origin $\left( {0,0} \right)$ go to its right side horizontally and mark 1 unit on the x-axis, and form 1 go upward up-to 2 units parallel to the y-axis.

**Complete step-by-step answer:**

Step 1: Plot the given coordinates

The points $A\left( {2.5,7.7} \right)$ and \[B\left( {3.7,5.3} \right)\]

Given that the x-axis is denoted by $\ln x$ and the y-axis is denoted by $\ln y$.

Step 2: Find the slope of the given straight line

The slope of a line is the tangent of the angle between the straight line and the positive direction of the x-axis. The slope is denoted by $m$.

Let coordinate of point \[P\left( {{x_1},{y_1}} \right)\] and \[Q\left( {{x_2},{y_2}} \right)\]

The slope of a line when coordinates of any two points on the line are given by:

Slope, $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

The slope of the given straight line with coordinates $A\left( {2.5,7.7} \right)$ and \[B\left( {3.7,5.3} \right)\] as \[A\left( {{x_1},{y_1}} \right)\] and \[B\left( {{x_2},{y_2}} \right)\] respectively.

Therefore, slope $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{5.3 - 7.7}}{{3.7 - 2.5}}$

$

m = \dfrac{{ - 2.4}}{{1.2}} \\

\because m = - 2 \\

$

Step 3: Solve (a)

The slope of the given line is -2, no matter which on lines are used to calculate, the slope will be the same.

Let the value of $\ln y$is t when $\ln x$ is 0.

Thus the slope of the given straight line with coordinates $A\left( {2.5,7.7} \right)$ and \[C\left( {0,t} \right)\] as \[A\left( {{x_1},{y_1}} \right)\] and \[C\left( {{x_2},{y_2}} \right)\] respectively.

Therefore, slope $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{t - 7.7}}{{0 - 2.5}}$

$

\Rightarrow - 2 = \dfrac{{t - 7.7}}{{ - 2.5}} \\

\Rightarrow t - 7.7 = \left( { - 2} \right)\left( { - 2.5} \right) = 5 \\

\Rightarrow t = 5 + 7.7 \\

\because t = 12.7 \\

$

Hence, $\ln y$= 12.7

**Final answer: The value of $\ln y$ is 12.7 when $\ln x$ is 0.**

**Note:**The value of $\ln y$can also be found by using the equation of the straight line.

The equation of a straight line with the slope, $m$of line and a point \[\left( {{x_1},{y_1}} \right)\]on the line, is given by:

\[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\]

The slope of the given straight line as calculated in step 2, is -2

$\because m = - 2$

Thus the equation of the given straight line with coordinates $A\left( {2.5,7.7} \right)$ as a point \[\left( {{x_1},{y_1}} \right)\] on the line is given by:

\[

\left( {y - 7.7} \right) = - 2\left( {x - 2.5} \right) \\

\Rightarrow y - 7.7 = - 2x + 5 \\

\Rightarrow y + 2x - 12.7 = 0 \\

\]

Let the value of $\ln y$is t when $\ln x$ is 0. Thus, the coordinates $\left( {0,t} \right)$ lie on the given straight line as well, therefore it will satisfy the equation of the line.

Thus \[y + 2x - 12.7 = 0\] at $\left( {0,t} \right)$

\[

t + 2\left( 0 \right) - 12.7 = 0 \\

\because t = 12.7 \\

\]

Hence, $\ln y$= 12.7

The x-axis and y-axis divide the Cartesian plane into four parts, each part is known as a quadrant.

Pictorial representation of quadrants:

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