The point (1,0) is common to all log functions of the form k(x) = logb x with any base (b). Logarithmic functions share this feature because of a property of exponents that states b0 = 1 for any value of b. In a log function of the form k(x) = logb x, k(1) = logb 1.

So, by the definition of a logarithm, bk(1) = 1.

Applying the property of exponents to bk(1) = 1 = b0, we get k(1) = 0.

So, (1,0) is a point on the graph.

Ibelieve that is true! but go with your gut.good luck!

The answer is    (d) 7/2======we have    we also  have an integral      notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum:     notice that for the piecewise function f, the function has the equationf(x) = x when x  ≤ 0. so for the integral  , f(x) = x so it changes into      for the piecewise function f, the function has the equationf(x) = x+1  when x  > 0. for the integral  , we have    therefore, back to our splitted integral:    

Option b is correct. the violation of society's formally enacted criminal law