# Solve for a and b in y = a logx + b with 2 known points

#### pencile

##### New member
Is it possible for a and b to be found if only points (x1,y1) and (x2,y2) are known?

y = a log(x) + b
x = 10( (y - b) / a )
xa = ( 10y / 10b )

I can't see a way to separate a and b from the variables.
Thanks.

#### JeffM

##### Elite Member
Is it possible for a and b to be found if only points (x1,y1) and (x2,y2) are known?

y = a log(x) + b
x = 10( (y - b) / a )
xa = ( 10y / 10b )

I can't see a way to separate a and b from the variables.
Thanks.
Let's think about the first problem.

$$\displaystyle y_1 = a log_{10}(x_1) + b \implies b = y_1 - a log_{10}(x_1).$$

$$\displaystyle y_2 = a log_{10}(x_2) + b = a log_{10}(x_2) + y_1 - a log_{10}(x_1) \implies a = (y_2 - y_1) \div log_{10} \left ( \dfrac{x_2}{x_1} \right ).$$

So yes you can find a and b in the first problem. Do you follow that? If so try the others on your own.

#### stapel

##### Super Moderator
Staff member
Is it possible for a and b to be found if only points (x1,y1) and (x2,y2) are known?

y = a log(x) + b
x = 10( (y - b) / a )
xa = ( 10y / 10b )

I can't see a way to separate a and b from the variables.
I'm not sure what you mean by "separating a and b from the variables"...? You've been given two points (that is, two values for each of x and y) and an equation that relates x and y. (The second and third equations in your post are just rearrangements of the first equation.) You've been asked to plug the two data points into the equation (in place of x and y), giving you two equations in two unknowns (a and b). Then you're asked to solve that system of equations.

This is quite solvable, and there is no need to "solve" the equation for "a" and "b" first. (And a good thing, since it isn't possible to find their values without plugging the given data points in!)

Please plug the given points into the given equation, and simplify to find the resulting system of equations. If you get stuck in solving the system, please reply showing all of your work so far. Thank you!