Question

Calculate $ {\log _2}32 + {\log _4}49 + {\log _8}125 - {\log _2}1120 $

Answer 1

Hint: Here the question is related to logarithmic terms. To solve this question, we have formula $ {\log _{{a^n}}}b = {\log _a}{(b)^{\dfrac{1}{n}}} $ . We first calculate the value for the logarithmic terms and then we add upon the terms and hence we obtain the desired results.

Complete step-by-step answer:
Consider the equation $ {\log _2}32 + {\log _4}49 + {\log _8}125 - {\log _2}1120 $ ---------------(1)
 Now we apply the formula $ {\log _{{a^n}}}b = {\log _a}{(b)^{\dfrac{1}{n}}} $ to each term. Now we find the value for each term so we have
Consider the first term $ {\log _2}32 $
By applying the formula, we have
 $ {\log _2}32 = {\log _{{{(2)}^1}}}32 = {\log _2}{(32)^1} = {\log _2}32 $
Therefore $ {\log _2}32 = {\log _2}32 $ ---------------(2)
Consider the second term $ {\log _4}49 $
By applying the formula, we have
4 an be written as 2 to the power of 2 i.e., $ 4 = {(2)^2} $
 $ {\log _4}49 = {\log _{{{(2)}^2}}}49 = {\log _2}{(49)^{\dfrac{1}{2}}} = {\log _2}\sqrt {49} $
The square root of 49 is 7 so $ {\log _2}\sqrt {49} $ can be written as $ {\log _2}7 $
Therefore $ {\log _4}49 = {\log _2}7 $ -----------------(3)
Consider the third term $ {\log _8}125 $
By applying the formula, we have
8 an be written as 2 to the power of 3 i.e., $ 8 = {(2)^3} $
 $ {\log _8}125 = {\log _{{{(2)}^3}}}125 = {\log _2}{(125)^{\dfrac{1}{3}}} = {\log _2}\sqrt[3]{{125}} $
The cube root of 125 is 5 so $ {\log _2}\sqrt[3]{{125}} $ can be written as $ {\log _2}5 $
Therefore $ {\log _8}125 = {\log _2}5 $ -----------------(4)
Consider the fourth term $ {\log _2}1120 $
By applying the formula, we have
 $ {\log _2}1120 = {\log _{{{(2)}^1}}}1120 = {\log _2}{(1120)^1} = {\log _2}1120 $
Therefore $ {\log _2}1120 = {\log _2}1120 $ --------------(5)
Substituting equation(2,) equation (3), equation (4) and equation(5) in equation(1)
Hence, we have
 $ {\log _2}32 + {\log _4}49 + {\log _8}125 - {\log _2}1120 = {\log _2}32 + {\log _2}7 + {\log _2}5 - {\log _2}1120 $
We apply the logarithmic property $ {\log _2}m + {\log _2}n = {\log _2}(m \times n) $
So, we have
 $ \Rightarrow {\log _2}(32 \times 7 \times 5) - {\log _2}1120 $
On simplification we have
 $
   \Rightarrow {\log _2}(224 \times 5) - {\log _2}1120 \\
   \Rightarrow {\log _2}1120 - {\log _2}1120 \\
  $
For the above inequality we apply the property $ {\log _2}m - {\log _2}n = {\log _2}\left( {\dfrac{m}{n}} \right) $
So we have
 $
   \Rightarrow {\log _2}\left( {\dfrac{{1120}}{{1120}}} \right) \\
   \Rightarrow {\log _2}1 \\
  $
The value of $ {\log _2}1 $ is zero
Therefore
 $ \Rightarrow {\log _2}1 = 0 $
Hence, we have $ {\log _2}32 + {\log _2}49 + {\log _8}125 - {\log _2}1120 = 0 $
So, the correct answer is “0”.

Note: The question contains the log terms we must know the logarithmic properties which are the standard properties. By applying properties we can solve the question in an easy manner. The base of log is a power of number then it has a specified formula that is $ {\log _{{a^n}}}b = {\log _a}{(b)^{\dfrac{1}{n}}} $ . We apply the formula where it is necessary. Hence, we obtain the desired result.