The point of "logarithmic graphs" is that instead of graphing "V" against "Y" you graph "log(V)" against log(Y). Any equation of the form becomes, after, taking the logarithm of both sides, , which has astraight lineas graph (of log(V) as a function of log(Y)). And straight lines are easier to recognize than other curves.

Now if Y= 0.52 and V= 2.88, then log(Y)= -0.2840 and log(V)= 0.4594.

If Y= 0.73 and V= 2.05, then log(Y)= -0.1367 and log(V)= 0.3118.

If Y= 0.94 and V= 1.60, then log(Y)= -0.02687 and log(V)= 0.2041.

If Y= 1.23 and V= 1.22, then log(Y)= 0.2041 and log(V)= 0.001507.

If Y= 1.57 and V= 0.96 then log(Y)= 0.1959 and log(V)= -0.01773.

Graph those (log(Y), log(V)) points. I haven't graphed them myself but I suspect that the whole point of this problem is that they lie, a least roughly, along a straight line. A straight line can be written "y= ax+ b". There are a variety of ways of finding "a" and "b"- "a" itself is the "slope" of the line and "b" the "y-intercept" but any two points will give two equations that can be solved for a and b. If you were given a problem like this you are clearly expected to know how to find those coefficients for a straight line.

Since here, your "x" is "log(Y)" and "y" is "log(V)", "y= ax+ b" becomes "log(V)= alog(Y)+ b". Take 10 (I calculated "common logs" above- if I had used the "natural logarithm I would use "e") to the power of each side, we have or .