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continuing from above... = f(a) + f(a^{n}) = f(a) + n\cdot f(a) = (n+1) \cdot f(a)

Let b = \frac{1}{b}, then substituting into f(ab) = f(a) + f(b): f(a\cdot\frac{1}{b}) = f(a) + f(\frac{1}{b}) From part (b) it follows that f(\frac{1}{b}) = -f(b), therefore f(a\cdot\frac{1}{b}) = f(a) - f(b)

From part a: If ab = 1, then f(1) = 0 and b = 1/a Substituting these values into f(ab) = f(a) + f(b) 0 = f(a) + f(1/a) f(1/a) = -f(a)

If b = 1 then f(a) = f(a) + f(1) therefore f(1) = 0. Thanks

You are correct - fixed it in original post. Thank you.

Suppose that we have a function f(x) such that f(ab) = f(a)+f(b) for all rational numbers a and b. (a) Show that f(1) = 0. (b) Show that f(1/a) = -f(a). (c) Show that f(a/b) = f(a) - f(b). (d) Show that f(an) = nf(a) for every positive integer a. For (a), if ab = 1 then a = 1/b and b = 1/a. Not...

This implies that ab = 1

I have tried various methods to solve this... If logb(a) = loga(b) where a != b (!= means does not equal), ab > 0 and neither a nor b are 1, then what is the value of ab?

I contacted the author of the book. To my surprise I got a quick response. He stated that the problem is not solvable without saying that EFGA is a parallelogram. There was an error during the printing process and the problem was supposed to state this fact. Problem solved!

I spent days on this one to no avail. I can pick an E such that AE is parallel to GF, but that does not allow me to show that AE = GF. I really believe the author meant EFGA not to be initially defined as a parallelogram. I sure do wish I could see their solution to this one. Thanks for the reply!

That is true if you assume quadrilateral EFGA is a parallelogram. Only quadrilateral ABCD is defined as a parallelogram. You would have to prove that EFGA is a parallelogram, then the problem becomes easy using area formulas. How do you prove that EFGA is a parallelogram with the given information?

Using the theorems: Area of a triangle = (1/2)bh Area of a parallelogram = bh Area of a trapezoid = (1/2)h(b1+b2) Solve the following: