Question

Show that for all real numbers aand b with a>1 and b>1, if f(x) is O (log _bx), then f(x) is O (log _ax).

Short answer

Hence we conclude f(x) is O (log _b x), then f(x) is O (log _a x)

Step-by-step solution

Step 1

Given, f(x) is O (log _b x)

By the definition of Big-O notation, there exist a positive real number M∋,

│f(x)│≤M│g(x)│, whenever x>k

Step 2

Consider,

│f(x)│≤M │log _b x │

=M$\left|\frac{{\log }_{a}x}{{\log }_{b}x}\right|$

=$=\frac{M}{{\log }_{b}x}\left|\log ax\right|$

Step 3

Let M₁=$\frac{M}{{\log }_{b}x}$, then

│f(x)│≤M₁ │log _a x │ whenever x>k

Therefore by the definition of Big-O notation, f(x) is O (log _a x) with constant M and k.

Final answer

Hence we conclude f(x) is O (log _b x), then f(x) is O (log _a x)