Question

Solving $ 1+{{\log }_{2}}\left( x-1 \right)\le {{\log }_{x-1}}4 $ we get domain of $ x $ be $ \left( m,n \right] $ .Find $ m+n $ . \[\]

Answer 1

Hint: We use the fact that the domain of logarithmic function is positive real number set to find $ m $ . We use logarithmic identities to make the function $ f\left( x \right)={{\log }_{2}}2\left( x-1 \right)-{{\log }_{x-1}}4 $ . We find for what $ x $ the function $ f\left( x \right)=0 $ and then use wavy curve method to find $ n $ . \[\]


Complete step by step answer:
We know that the logarithm is the inverse operation to exponentiation. If $ {{b}^{y}}=x $ then the logarithm denoted as log and calculated as with base $ b $ and argument $ x $ as
\[{{\log }_{b}}x=y\]
Here $ x,y,b $ are real numbers subjected to condition $ x>0 $ and $ b>0,b\ne 1 $ .We know the logarithmic identity involving power $ m\ne 0 $ where $ m $ is real number as
\[m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}\]
We also know the logarithmic identity involving product as
\[{{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n\]
 We are given
 $ 1+{{\log }_{2}}\left( x-1 \right)\le {{\log }_{x-1}}4 $
We use the logarithmic identity $ {{\log }_{b}}b=1 $ for $ b=2 $ to have;
\[{{\log }_{2}}2+{{\log }_{2}}\left( x-1 \right)\le {{\log }_{x-1}}4\]

We use the logarithmic identity of product for $ m=2,n=x-1,b=2 $ in the left hand side of the above step to have;
\[\begin{align}
  & {{\log }_{2}}2\left( x-1 \right)\le {{\log }_{x-1}}4 \\
 & \Rightarrow {{\log }_{2}}2\left( x-1 \right)-{{\log }_{x-1}}4\le 0 \\
\end{align}\]
Let us define a function such that
\[f\left( x \right)={{\log }_{2}}2\left( x-1 \right)-{{\log }_{x-1}}4\]
We see that since logarithmic functions can take only positive real numbers as arguments we have $ 2\left( x-1 \right)>0\Rightarrow x>1 $ . Since the base cannot be negative or 1 we have again $ x-1>0\Rightarrow x>1 $ .So the minimum value for which $ f\left( x \right) $ is defined is 1 so we have $ m=1 $ . We need to find the zeroes of $ f\left( x \right) $ . So let us consider
\[f\left( x \right)={{\log }_{2}}2\left( x-1 \right)-{{\log }_{x-1}}4=0\]
We use trial and error methods to intuitively guess $ f\left( x \right)=0 $ when $ x-1=2 $ . So we have;
\[\begin{align}
  & x-1=2 \\
 & \Rightarrow x=3 \\
\end{align}\]
Since $ {{\log }_{2}}2\left( x-1 \right) $ and $ {{\log }_{x-1}}4 $ are strictly increasing curves they can intersect at most at one point and hence $ f\left( x \right) $ has only one zero and that is 3. So function $ f\left( x \right) $ will change the sign at $ x=3 $ . So let us draw the wavy curves


So the maximum value for which $ f\left( x \right)\le 0 $ is $ x=3 $ and hence $ n=3 $ . So we have
\[\begin{align}
  & \left( m,n \right]=\left( 1,3 \right] \\
 & \Rightarrow m+n=1+3=4 \\
\end{align}\]

Note:
We can easily check that $ f\left( x \right)>0 $ for all $ x > 3 $ by putting $ x=5 $ in $ f\left( x \right) $ . We call a function strictly increasing when $ f\left( {{x}_{1}} \right) < f\left( {{x}_{2}} \right) $ implies $ {{x}_{1}} < {{x}_{2}} $ for all $ {{x}_{1}},{{x}_{2}} $ in the domain of $ f\left( x \right) $ . We call function monotonically increasing when $ f\left( {{x}_{1}} \right)\le f\left( {{x}_{2}} \right) $ implies $ {{x}_{1}}\le {{x}_{2}} $ .