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Question

Question: How do you write the following expression as a single logarithm: \( \log 5 - \log x - \log y \) ?...

How do you write the following expression as a single logarithm: log5logxlogy\log 5 - \log x - \log y ?

Explanation

Solution

Hint : The given function is the logarithm function it can be defined as logarithmic functions are the inverses of exponential functions. By using one of the Basic Properties of logarithmic that is log(mn)=logmlogn\log \left( {\dfrac{m}{n}} \right) = \log m - \log n we can write the given function as a single logarithm function.

Complete step-by-step answer :
The function from positive real numbers to real numbers to real numbers is defined as logb:R+Rlogb(x)=y{\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y , if by=x{b^y} = x , is called logarithmic function or the logarithm function is the inverse form of exponential function.
There are some basic logarithms properties
1. product rule :- log(mn)=logm+logn\log \left( {mn} \right) = \log m + \log n
2. Quotient rule :- log(mn)=logmlogn\log \left( {\dfrac{m}{n}} \right) = \log m - \log n
3. Power rule :- log(mn)=n.logm\log \left( {{m^n}} \right) = n.\log m
Here in the given function solve by using a property of quotient rule of logarithm function
Defined as log(mn)=logmlogn\log \left( {\dfrac{m}{n}} \right) = \log m - \log n
Now, Consider the logarithm function
log5logxlogy\Rightarrow \,\log 5 - \log x - \log y (1)
Then, equation (1) can be rewritten as
(log5logx)logy\Rightarrow \,\,\left( {\log 5 - \log x} \right) - \log y (2)
Using quotient rule of logarithm function to the function log5logx\log 5 - \log x where M=log5M = \log 5 and n=logxn = \log x , then log(mn)=log(5x)\Rightarrow \,\,\log \left( {\dfrac{m}{n}} \right) = \log \left( {\dfrac{5}{x}} \right)
Then, equation (2) becomes
log(5x)logy\Rightarrow \,\,\log \left( {\dfrac{5}{x}} \right) - \log y (3)
Again using the quotient rule of logarithm function to the function log(5x)logy\log \left( {\dfrac{5}{x}} \right) - \log y where m=log(5x)m = \log \left( {\dfrac{5}{x}} \right) and n=logyn = \log y , then log(mn)\log \left( {\dfrac{m}{n}} \right) of equation (3) becomes
log(5xy)\Rightarrow \,\,\,\log \left( {\dfrac{{\dfrac{5}{x}}}{y}} \right)
log(5x.1y)\Rightarrow \,\,\,\log \left( {\dfrac{5}{x}.\dfrac{1}{y}} \right)
log(5xy)\Rightarrow \,\,\,\log \left( {\dfrac{5}{{xy}}} \right)
The single logarithm expression of log5logxlogy\log 5 - \log x - \log y is log(5xy)\,\,\,\log \left( {\dfrac{5}{{xy}}} \right)
Verification:
By verify the given solution consider the single logarithm function log(5xy)\,\,\log \left( {\dfrac{5}{{xy}}} \right) by using product and quotient rule we can verify
Consider log(5xy)\,\,\log \left( {\dfrac{5}{{xy}}} \right) (4)
Where m=5m = 5 and n=xyn = xy
By using quotient rule of differentiation to equation (4) can be written as
log5logxy\Rightarrow \,\,\log 5 - \log xy (5)
Apply product rule of logarithm function to the function logxy\log xy , where m=xm = x and n=yn = y , then
logxy=logx+logy\log xy = \log x + \log y
Then equation (5) can be written as
log5(logx+logy)\Rightarrow \,\,\log 5 - \left( {\log x + \log y} \right)
log5logxlogy\Rightarrow \,\,\log 5 - \log x - \log y
Hence it verified
So, the correct answer is “ log5logxlogy\log 5 - \log x - \log y is log(5xy)\,\,\,\log \left( {\dfrac{5}{{xy}}} \right) ”.

Note : If the function contains the log term, then the function is known as logarithmic function. We have two types of logarithms namely common logarithm and natural logarithm. Since it involves the arithmetic operations, we have a standard logarithmic property for the arithmetic operations. By using the properties, we can solve these types of questions.