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Question: If \( {h^2} + {k^2} = 23hk \) where \( h > 0 \) and \( k > 0 \) . Show that \( \log \left( {\dfrac{{...

If h2+k2=23hk{h^2} + {k^2} = 23hk where h>0h > 0 and k>0k > 0 . Show that log⁡(h+k5)=12(log⁡h+log⁡k)\log \left( {\dfrac{{h + k}}{5}} \right) = \dfrac{1}{2}\left( {\log h + \log k} \right) ?

Explanation

Solution

Hint : In the above question, we are given an equation in two variables, h and k and we have to prove a given equation that it holds true. The given equation involves logarithmic functions in both variables which gives us an idea to take log on both sides of the equation given to us. So to solve this question, we must know what logarithm functions are. A logarithm function is the inverse of an exponential function. To do this question, we will use the laws of the logarithm.

Complete step-by-step answer :
In the question, we are given an equation h2+k2=23hk{h^2} + {k^2} = 23hk .
We add 2hk2hk on both sides of the equation so as to make the left side of the equation as square of a binomial.
⇒h2+k2+2hk=23hk+2hk\Rightarrow {h^2} + {k^2} + 2hk = 23hk + 2hk
⇒h2+k2+2hk=25hk\Rightarrow {h^2} + {k^2} + 2hk = 25hk
Condensing the left side of the equation into Whole Square, we get,
⇒(h+k)2=25hk\Rightarrow {\left( {h + k} \right)^2} = 25hk
Now, taking log on both sides of the equation, we get,
⇒log⁡(h+k)2=log⁡(25hk)\Rightarrow \log {\left( {h + k} \right)^2} = \log \left( {25hk} \right)
Now using the property of logarithm log⁡xn=nlog⁡(x)\log {x^n} = n\log \left( x \right) ,
⇒2log⁡(h+k)=log⁡(25hk)\Rightarrow 2\log \left( {h + k} \right) = \log \left( {25hk} \right)
Using logarithmic property, log⁡ax+log⁡ay=log⁡a(xy){\log _a}x + {\log _a}y = {\log _a}\left( {xy} \right)
⇒2log⁡(h+k)=log⁡(25)+log⁡h+log⁡k\Rightarrow 2\log \left( {h + k} \right) = \log \left( {25} \right) + \log h + \log k
Now, again using the property of logarithm log⁡xn=nlog⁡(x)\log {x^n} = n\log \left( x \right) ,
⇒2log⁡(h+k)=2log⁡(5)+log⁡h+log⁡k\Rightarrow 2\log \left( {h + k} \right) = 2\log \left( 5 \right) + \log h + \log k
⇒2log⁡(h+k)−2log⁡(5)=log⁡h+log⁡k\Rightarrow 2\log \left( {h + k} \right) - 2\log \left( 5 \right) = \log h + \log k
Using logarithmic property log⁡ax−log⁡ay=log⁡a(xy){\log _a}x - {\log _a}y = {\log _a}\left( {\dfrac{x}{y}} \right) ,
⇒2log⁡(h+k5)=log⁡h+log⁡k\Rightarrow 2\log \left( {\dfrac{{h + k}}{5}} \right) = \log h + \log k
⇒log⁡(h+k5)=(log⁡h+log⁡k2)\Rightarrow \log \left( {\dfrac{{h + k}}{5}} \right) = \left( {\dfrac{{\log h + \log k}}{2}} \right)
Hence, Proved.

Note : There are several laws of the logarithm that make the calculations easier and help us evaluate the logarithm functions. The standard base of logarithm functions is 10, that is, if we are given a function without any base like log⁡x\log x then we take the base as 10. Now while applying the laws of the logarithm, we should keep in mind an important rule that is the base of the logarithm functions involved should be the same in all the calculations, as the base of both the functions in the question is the same, we can apply the logarithm laws in the given question.