Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same order. (Note: Consider to prove the result, first prove f(x) = O(g(x)) and then prove g(x) = O(f(x)). f(x) = log (x² +1) and g(x) = logx. Reset Step 1 Step 2 Step 3 Step 4 Since the two functions are positive for x>0, we may insert absolute values without changing the quantities to get Ig(x)| ≤ If(x) for x>0. This proves that fis big-O of g. First we have to prove that fis big-O of g, i.e. that If(x)I s Clg(x)l for all x> k, where C and k are constants. Observe that for all x> 1, x +1 < x²+x² = 2x². By applying the log to that inequality and using laws of logarithms, we get that log(x + 1) < log(2x) = log2 + 2logx First we have to prove that fis big-O of g, i.e. that If(x)l z Clg(x)l for all x> k, where C and k are constants. Observe that for all x>0, x² + 1 < x² + x² = 2x². By applying the log to that inequality and using laws of logarithms, we get that log(x² + 1) < log(2x²) = log2 + 2logx. Under the stricter assumption x>2, log2 1, log(x + 1) is positive, hence we can apply absolute values to both side of the inequality without changing it and get llog(x² + 1)| ≤ 3llogxl. We have shown that fis big-O of g. Under the stricter assumption x>2, log2 < logx, hence log(x + 1) ≤ 2logx. Since x² + 1 > 1, log(x² + 1) is positive, hence we can apply absolute values to both side of the inequality without changing it and get llog(x² + 1)I s2llogxl. We have shown that fis big-O of g. Secondly, we have to prove that g is big-O of f, i.e. that Ig(x)l z Cif(x)l for all x>k, where C and k are constants. For x>0, x k, where C and k are constants. For x>0, x 1, we may insert absolute values without changing the quantities to get Ig(x)I sIf(x) for x>1. This proves that g is big-O of f. Having both proved that fis big-O of g, and g is big-O of f, we have shown that fand g have the same order.