Solveeit Logo
Login

Question

Question: How do you simplify \[\dfrac{{\ln \left( {\dfrac{1}{4}} \right)}}{{\ln 2}}\]?...

How do you simplify ln⁡(14)ln⁡2\dfrac{{\ln \left( {\dfrac{1}{4}} \right)}}{{\ln 2}}?

Explanation

Solution

When one term is raised to the power of another term, the function is called an exponential function, for example a=xya = {x^y} . The inverse of the exponential functions are called logarithm functions, the inverse of the function given in the example isy=log⁡xay = {\log _x}a that is a logarithm function. These functions also obey certain rules called laws of the logarithm, using these laws we can write the function in a variety of ways. Using the laws of the logarithm, we can solve the given problem.

Complete step by step solution:
Given,
ln⁡(14)ln⁡2\dfrac{{\ln \left( {\dfrac{1}{4}} \right)}}{{\ln 2}}.
We know the logarithm quotient rule that the logarithm of the ratio of two quantities is the
logarithm of the numerator minus the logarithm of the denominator.
That is ln⁡(mn)=ln⁡m−ln⁡n\ln \left( {\dfrac{m}{n}} \right) = \ln m - \ln n. Applying this in the numerator term we have,
ln⁡(14)ln⁡2=ln⁡(1)−ln⁡(4)ln⁡2\dfrac{{\ln \left( {\dfrac{1}{4}} \right)}}{{\ln 2}} = \dfrac{{\ln (1) - \ln (4)}}{{\ln 2}}
We know that ln⁡(1)=0\ln (1) = 0.
=0−ln⁡(4)ln⁡(2)= \dfrac{{0 - \ln (4)}}{{\ln (2)}}
=−ln⁡(4)ln⁡(2)= - \dfrac{{\ln (4)}}{{\ln (2)}}
But we can write 4=224 = {2^2}then
=−ln⁡(22)ln⁡2= - \dfrac{{\ln ({2^2})}}{{\ln 2}}
We know the logarithm power rule that the logarithm of an exponential number is the exponent times the logarithm of the base.
That is ln⁡(mn)=nln⁡(m)\ln ({m^n}) = n\ln (m). Applying this we have
=−2ln⁡(2)ln⁡(2)= - \dfrac{{2\ln (2)}}{{\ln (2)}}
=−2= - 2
Thus we have
⇒ln⁡(14)ln⁡2=−2\Rightarrow \dfrac{{\ln \left( {\dfrac{1}{4}} \right)}}{{\ln 2}} = - 2

Note: the above problem can be solved easily by only using the power rule,
ln⁡(14)ln⁡2\dfrac{{\ln \left( {\dfrac{1}{4}} \right)}}{{\ln 2}},
Since we can write 14\dfrac{1}{4} as 122\dfrac{1}{{{2^2}}}
=ln⁡(122)ln⁡2= \dfrac{{\ln \left( {\dfrac{1}{{{2^2}}}} \right)}}{{\ln 2}}
Rearranging we have
=ln⁡(2−2)ln⁡2= \dfrac{{\ln \left( {{2^{ - 2}}} \right)}}{{\ln 2}}
We know the logarithm power rule that is ln⁡(mn)=nln⁡(m)\ln ({m^n}) = n\ln (m). Applying this we have
=−2ln⁡(2)ln⁡2= \dfrac{{ - 2\ln \left( 2 \right)}}{{\ln 2}}
=−2= - 2. In both cases we have the same answer.
The important condition while applying the laws of the logarithm is that the base of the logarithm functions involved should be the same in all the calculations. Here we have a natural logarithm function that is base′e′'e'. If they simply given log⁡x\log x then we take the standard base that is 10.