If point c divides point ab in the ratio 2: 3 the coordinates of c are . if point d divides point ac in the ratio 3: 2 the coordinates of d are

check the picture below

for point C

\bf \textit{internal division of a line segment}
\\ \quad A(1,4)\qquad B(6,-1)\qquad
ratio1=2\qquad ratio2=3\qquad 2:3

\cfrac{A{ C }}{{ C }B}=\cfrac{ratio1}{ratio2}\implies 
ratio2\cdot A=ratio1\cdot B\quad \implies 
3(1,4)=2(6,-1)
\\ \quad {{ C=\left(\cfrac{\textit{sum of "x" values}}{ratio1+ratio2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{ratio1+ratio2}\right)}}\\ \quad \\

\bf thusC=\left(\cfrac{(3\cdot 1)+(2\cdot 6)}{2+3}\quad ,\quad \cfrac{(3\cdot 4)+(2\cdot -1)}{2+3}\right)

and for point D

\bf \textit{internal division of a line segment}
\\ \quad 
A(1,4)\qquad B(6,-1)\qquad
ratio1=3\qquad ratio2=2\qquad 3:2

\cfrac{A{ D }}{{ D }B}=\cfrac{ratio1}{ratio2}\implies 
ratio2\cdot A=ratio1\cdot B\quad \implies 
2(1,4)=3(6,-1)
\\ \quad 
{{ D=\left(\cfrac{\textit{sum of "x" values}}{ratio1+ratio2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{ratio1+ratio2}\right)}}

\bf thus

D=\left(\cfrac{(2\cdot 1)+(3\cdot 6)}{3+2}\quad ,\quad \cfrac{(2\cdot 4)+(3\cdot -1)}{3+2}\right)

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.