
If x = log 0.6, y = log 1.25 and z = log 3 – 2 log 2, then find the value of x + y – z.
A.0
B.1
C.2
D.3
Answer
604.8k+ views
Hint: We need to know the basic logarithmic formulae to solve the given problem.
Use product and division rule to solve the given problem.
Given x = log 0.6
y = log 1.25
z = log 3 – 2 log 2
x + y – z = log 0.6 + log 1.25 – (log 3 – 2 log 2)
$\left[ {\because \log A + \log B = \log (A \times B)\;\& \;\log A - \log B = \log \dfrac{A}{B}} \right]$
x + y – z = log (0.6$ \times $1.25) – log$\left( {\dfrac{3}{4}} \right)$
x + y – z = log 0.75 – log 0.75 = 0
$\therefore $ The value of x + y – z = 0
Note:
Logarithm is a way of showing how big a number is in terms of how many times you have to multiply a certain number (the base) to get it. It is the inverse function of exponentiation.
$ \Rightarrow y = {\log _e}x$ the inverse of this function will be $y = {e^x}$. We used product rule $\log A + \log B = \log (A \times B)$ and division rule $\log A - \log B = \log \dfrac{A}{B}$ in the above problem.
Use product and division rule to solve the given problem.
Given x = log 0.6
y = log 1.25
z = log 3 – 2 log 2
x + y – z = log 0.6 + log 1.25 – (log 3 – 2 log 2)
$\left[ {\because \log A + \log B = \log (A \times B)\;\& \;\log A - \log B = \log \dfrac{A}{B}} \right]$
x + y – z = log (0.6$ \times $1.25) – log$\left( {\dfrac{3}{4}} \right)$
x + y – z = log 0.75 – log 0.75 = 0
$\therefore $ The value of x + y – z = 0
Note:
Logarithm is a way of showing how big a number is in terms of how many times you have to multiply a certain number (the base) to get it. It is the inverse function of exponentiation.
$ \Rightarrow y = {\log _e}x$ the inverse of this function will be $y = {e^x}$. We used product rule $\log A + \log B = \log (A \times B)$ and division rule $\log A - \log B = \log \dfrac{A}{B}$ in the above problem.
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