The title of the question "HOW TO PROVE THAT......." IS WRONG. If one has to prove the given equation, one must show that the equation is true for all values of x. But, for x= 1, the LHS (Left Hand ...

P dilip_k Mar 5, 2016 The given equation can be written as \displaystyle{1}+{2}{{\log}_{{x}}{\left({x}-{4}\right)}}={{\log}_{{x}}{\left(\frac{{{x}-{6}}}{{6}}\right)}}+{{\log}_{{x}}{\left({x}-{6}\right)}} ...

\displaystyle{x}^{{3}}+{8}{x}^{{2}}+{10}{x}+{3}={0} Explanation: \displaystyle{\left(\frac{{{\log}_{{2}}{\left({2}{x}-{3}\right)}}}{{{\log}_{{2}}{x}}}\right)}-{{\log}_{{x}}{\left({x}+{6}\right)}}+{\left(\frac{{1}}{{{\log}_{{{x}+{2}}}{x}}}\right)}={1} ...

\log_3(3x + 2) = \log_9(4x + 5) Convert the RHS to base 3 to get \log_3 (3x+2) = \frac{\log_3 (4x+5)}{\log_3 3^2} So that you get 2\log_3 (3x+2) = \log_3 (4x+5) The power law for ...

You want \frac{2.2}{2^n} < 10^{-2}, or in other words, 220 < 2^n. Then \log_2 220 < n. Since \log_2 220 \approx 7.7814\cdots, we see that the smallest such n is n=8. Alternatively, you ...

Your proofs are fine. In a sense, both inequalities you proved are 'special cases' of the general inequality you started with. Since they are different special cases, it makes sense that you need to ...