Question

Prove that $ 7\log \dfrac{{16}}{{15}} + 5\log \dfrac{{25}}{{24}} + 3\log \dfrac{{81}}{{80}} = \log 2 $

Answer 1

Hint: In order to prove the above equation , first take the left-hand side of the equation. Simplify the LHS by first separating all the terms using rule $ {\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n) $ . Now open the brackets and combine all the terms, and you will get the answer which is equal to RHS of the equation. Hence proved.

Complete step-by-step answer:
To Prove: $ 7\log \dfrac{{16}}{{15}} + 5\log \dfrac{{25}}{{24}} + 3\log \dfrac{{81}}{{80}} = \log 2 $
Proof:
In order to prove the above logarithmic expression,
Taking Left-Hand side (LHS)
 $ LHS = 7\log \dfrac{{16}}{{15}} + 5\log \dfrac{{25}}{{24}} + 3\log \dfrac{{81}}{{80}} $
To simplify the above expression, we must know the properties of logarithms and with the help of them we are going to rewriting the above expression
Using the property of logarithm $ {\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n) $ in all the three terms we get
 $ = 7\left( {\log 16 - \log 15} \right) + 5\left( {\log 25 - \log 24} \right) + 3\left( {\log 81 - \log 80} \right) $
 $
   = 7\left( {\log {2^4} - \log \left( {5 \times 3} \right)} \right) + 5\left( {\log {5^2} - \log \left( {8 \times 3} \right)} \right) + 3\left( {\log {9^2} - \log \left( {16 \times 5} \right)} \right) \\
   = 7\left( {4\log 2 - \log 3 - \log 5} \right) + 5\left( {2\log 5 - \log {2^3} - \log 3} \right) + 3\left( {4\log 3 - \log {2^4} - \log 5} \right) \\
   = 28\log 2 - 7\log 3 - 7\log 5 + 10\log 5 - 15\log 2 - 5\log 3 + 12\log 3 - 12\log 2 - 3\log 5 \\
  $
Now combining all the like terms , we get
 $
   = 28\log 2 - 15\log 2 - 12\log 2 - 7\log 3 - 5\log 3 + 12\log 3 - 7\log 5 + 10\log 5 - 3\log 5 \\
   = 28\log 2 - 27\log 2 - 12\log 3 + 12\log 3 + 10\log 5 - 10\log 5 \\
   = \log 2 \\
  $
 $ LHS = \log 2 $
Now Taking Right-hand side
 $ RHS = \log 2 $
 $ LHS = RHS $
Hence proved

Note: 2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
 $ {\log _b}(mn) = {\log _b}(m) + {\log _b}(n) $
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
 $ {\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n) $
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
 $ n\log m = \log {m^n} $
6. The above guidelines work just if the bases are the equivalent. For example, the expression $ {\log _d}(m) + {\log _b}(n) $ can't be improved, on the grounds that the bases (the "d" and the "b") are not the equivalent, similarly as x2 × y3 can't be disentangled on the grounds that the bases (the x and y) are not the equivalent.