Question

If $ {\log _x}x - {\log _x}y = a $ , $ {\log _x}y - {\log _x}z = b $ and $ {\log _x}z - {\log _x}x = c $ , then find the value $ {\left( {\dfrac{x}{y}} \right)^{b - c}} \times {\left( {\dfrac{y}{z}} \right)^{c - a}} \times {\left( {\dfrac{z}{x}} \right)^{a - b}} $ .

Answer 1

Hint: Before attempting this question one should have prior knowledge of logarithm identities and also remember to use the log identity i.e. log a – log b = log ( $ \dfrac{a}{b} $ ), use this information to approach the solution to the question.

Complete step-by-step answer:
According to the given information we have $ {\log _x}x - {\log _x}y = a $
Since we know that by the logarithm identities i.e. $ \log a - \log b = \log \dfrac{a}{b} $
Using the above identities in $ {\log _x}x - {\log _x}y = a $ we get
 $ {\log _x}\left( {\dfrac{x}{y}} \right) = a $
Now by the identity of logarithm i.e. $ {\log _b}a = c $ $ \Rightarrow $ $ a = {b^c} $
  $ {\log _x}\left( {\dfrac{x}{y}} \right) = a $ $ \Rightarrow $ \[\dfrac{x}{y} = {x^a}\]
Now for $ {\log _x}y - {\log _x}z = b $
Since we know that by the logarithm identities i.e. $ \log a - \log b = \log \dfrac{a}{b} $
Using the above identities in $ {\log _x}y - {\log _x}z = b $ we get
 $ {\log _x}\left( {\dfrac{y}{z}} \right) = b $
Now by the identity of logarithm i.e. $ {\log _b}a = c $ $ \Rightarrow $ $ a = {b^c} $
  $ {\log _x}\left( {\dfrac{y}{z}} \right) = b $ $ \Rightarrow $ \[\dfrac{y}{z} = {x^b}\]
Now for $ {\log _x}z - {\log _x}x = c $
Since we know that by the logarithm identities i.e. $ \log a - \log b = \log \dfrac{a}{b} $
Using the above identities in $ {\log _x}z - {\log _x}x = c $ we get
 $ {\log _x}\left( {\dfrac{z}{x}} \right) = c $
Now by the identity of logarithm i.e. $ {\log _b}a = c $ $ \Rightarrow $ $ a = {b^c} $
  $ {\log _x}\left( {\dfrac{z}{x}} \right) = c $ $ \Rightarrow $ \[\dfrac{z}{x} = {x^c}\]
Since we have to find the value of $ {\left( {\dfrac{x}{y}} \right)^{b - c}} \times {\left( {\dfrac{y}{z}} \right)^{c - a}} \times {\left( {\dfrac{z}{x}} \right)^{a - b}} $
Substituting the values in the above equation we get
\[{\left( {{x^a}} \right)^{b - c}} \times {\left( {{x^b}} \right)^{c - a}} \times {\left( {{x^c}} \right)^{a - b}}\]
\[ \Rightarrow \]\[{x^a}^{b - ac} \times {x^b}^{c - ba} \times {x^c}^{a - cb}\]
\[ \Rightarrow \]\[{x^a}^{b - ac + bc - ab + ac - bc}\]
\[ \Rightarrow \]\[{x^0}\] = 1
Hence the value of $ {\left( {\dfrac{x}{y}} \right)^{b - c}} \times {\left( {\dfrac{y}{z}} \right)^{c - a}} \times {\left( {\dfrac{z}{x}} \right)^{a - b}} $ is equal to 1.

Note: In the above solution we simplified using the identities of logarithms which have some properties like product rule according to which the product of log is equal to the sum of the log with first base and with the second base i.e. $ {\log _x}\left( {ab} \right) = {\log _x}a + {\log _x}b $ , quotient rule according to which log is the quotient of a log is equal to the difference between the numerator and denominator log i.e. $ {\log _x}\dfrac{a}{b} = {\log _x}a - {\log _x}b $ and the rule of power which says that the new base of a log is equal to the log with new base which is divided by the log with old base i.e. $ {\log _b}x = \dfrac{{{{\log }_a}x}}{{{{\log }_a}b}} $ .