Question

Given ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. Prove that ${{\log }_{b+c}}a+{{\log }_{c-b}}a=2{{\log }_{b+c}}a{{\log }_{c-b}}a$ for all a > 0, a $\ne$ 1.

Answer 1

Consider the L.H.S of the term which we need to prove and apply the basic change rule given as ${{\log }_{n}}m=\dfrac{1}{{{\log }_{m}}n}$ to change the two logarithmic terms. Now, take L.C.M of the two terms and in the numerator apply the formula of logarithm given as, ${{\log }_{x}}p+{{\log }_{x}}q={{\log }_{x}}\left( pq \right)$ to simplify. Use the given relation: - ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$ and again apply the base change rule to take the denominator terms to the numerator and get the answer.

Complete step-by-step solution
Here, we have been provided a relation: - ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$ and we have to prove ${{\log }_{b+c}}a+{{\log }_{c-b}}a=2{{\log }_{b+c}}a{{\log }_{c-b}}a$.
Now, considering the L.H.S of the expression that we need to prove, we get,
$\Rightarrow$ L.H.S = ${{\log }_{b+c}}a+{{\log }_{c-b}}a$
Apply the base change rule of logarithm given as: - ${{\log }_{n}}m=\dfrac{1}{{{\log }_{m}}n}$, we get,
$\Rightarrow$ L.H.S = $\dfrac{1}{{{\log }_{a}}\left( b+c \right)}+\dfrac{1}{{{\log }_{a}}\left( c-b \right)}$
Taking L.C.M we get,
$\Rightarrow$ L.H.S = $\dfrac{{{\log }_{a}}\left( b+c \right)+{{\log }_{a}}\left( c-b \right)}{{{\log }_{a}}\left( b+c \right).{{\log }_{a}}\left( c-b \right)}$
Now, applying the formula: - ${{\log }_{x}}p+{{\log }_{x}}q={{\log }_{x}}\left( pq \right)$ in the numerator of the above expression, we get,
$\Rightarrow$ L.H.S = $\dfrac{{{\log }_{a}}\left[ \left( b+c \right).\left( c-b \right) \right]}{{{\log }_{a}}\left( b+c \right).{{\log }_{a}}\left( c-b \right)}$
Applying the algebraic identity: - $\left( c+b \right)\left( c-b \right)={{c}^{2}}-{{b}^{2}}$, we get,
$\Rightarrow$ L.H.S = $\dfrac{{{\log }_{a}}\left[ \left( {{c}^{2}}-{{b}^{2}} \right) \right]}{{{\log }_{a}}\left( b+c \right).{{\log }_{a}}\left( c-b \right)}$ - (1)
Here, we have been provided with the relation: - ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$, so we have,

& \Rightarrow {{a}^{2}}={{c}^{2}}-{{b}^{2}} \\\ & \Rightarrow {{c}^{2}}-{{b}^{2}}={{a}^{2}} \\\ \end{aligned}$$ Therefore, substituting the value of above relation in equation (1), we get, $$\Rightarrow $$ L.H.S = $$\dfrac{{{\log }_{a}}\left( {{a}^{2}} \right)}{{{\log }_{a}}\left( b+c \right).{{\log }_{a}}\left( c-b \right)}$$ Using the formula: - $${{\log }_{m}}\left( {{n}^{y}} \right)=y{{\log }_{m}}n$$, we get, $$\Rightarrow $$ L.H.S = $$\dfrac{2{{\log }_{a}}a}{{{\log }_{a}}\left( b+c \right).{{\log }_{a}}\left( c-b \right)}$$ Now, we know that if the base and argument of logarithm is the same then its value is 1. So, we have, $$\Rightarrow $$ L.H.S = $$\dfrac{2}{{{\log }_{a}}\left( b+c \right).{{\log }_{a}}\left( c-b \right)}$$ $$\Rightarrow $$ L.H.S = $$2\times \dfrac{1}{{{\log }_{a}}\left( b+c \right)}\times \dfrac{1}{{{\log }_{a}}\left( c-b \right)}$$ Again, applying the base change rule, we get, $$\Rightarrow $$ L.H.S = $$2\times {{\log }_{\left( b+c \right)}}a\times {{\log }_{\left( c-b \right)}}a$$ $$\Rightarrow $$ L.H.S = $$2{{\log }_{\left( b+c \right)}}a{{\log }_{\left( c-b \right)}}a$$ $$\Rightarrow $$ L.H.S = R.H.S Hence proved **Note:** One may note that we have been provided with a condition regarding the value of ‘a’ in the question, that is a > 0 and a $$\ne $$ 1. This condition is very important because it defines the logarithm. It says that the base of a logarithm cannot be negative, 0 or 1. Similarly, the argument cannot be negative or 0, however, it can be 1. You must remember all the important properties of the logarithm that are used above in the solution otherwise it will be very difficult to solve the question.