Question

1) Show that ${{\log }_{5}}6$ is irrational.
2) Find the least possible value of x such that ${{\log }_{\sin x}}\cos x+{{\log }_{\cos x}}\sin x=2$.

Answer 1

Hint: In this to show that ${{\log }_{5}}6$ is irrational we will first assume that ${{\log }_{5}}6$ is rational then it has the form $\dfrac{p}{q}$ then we will get contradiction after some steps that ${{\log }_{5}}6$ is not rational, which means it must be irrational. Then, for the second part, we will first use the property of logarithm that ${{\log }_{b}}a=\dfrac{\log a}{\log b}$. Then by cross multiplication, we will find some equation then by solving that equation, we get the least value of x.

Complete step-by-step solution
(1) Suppose ${{\log }_{5}}6$ is rational. Then
${{\log }_{5}}6=\dfrac{p}{q},\text{ }p,q\in {{Z }^{+}}\left( \text{positive integer } \right)$
${{5}^{{{\log }_{5}}6}}={{5}^{\dfrac{p}{q}}}$
By the property of logarithm, ${{a}^{{{\log }_{a}}b}}=b$
By applying this property, we get
$6={{5}^{\dfrac{p}{q}}}$
Taking q power on both sides, we get
\[{{6}^{q}}={{\left( {{5}^{\dfrac{p}{q}}} \right)}^{q}}\]
\[\Rightarrow {{6}^{q}}={{5}^{p}}......\left( i \right)\]
Since p and q are positive integers. Also, 6 is an even number and qth power of any even number is an even number. Similarly pth power of any odd number is an odd number.
\[{{6}^{q}}\]is an even number and \[{{5}^{p}}\] is an odd number. Hence equation (i) is a contradiction to that even number cannot be equal to an odd number. Hence our assumption is wrong.
Implies ${{\log }_{5}}6$ is an irrational number.
(2) Given that,
${{\log }_{\sin x}}\cos x+{{\log }_{\cos x}}\sin x=2$
By using the property ${{\log }_{b}}a=\dfrac{\log a}{\log b}$, we get
\[\dfrac{\log \cos x}{\log \sin x}+\dfrac{\log \sin x}{\log \cos x}=2\]
By cross multiplication, we get
\[{{\left( \log \cos x \right)}^{2}}+{{\left( \log \sin x \right)}^{2}}=2\log \sin x\cdot \log \cos x\]
By subtracting \[2\log \sin x\cdot \log \cos x\] by both sides, we get
\[\Rightarrow{{\left( \log \cos x \right)}^{2}}+{{\left( \log \sin x \right)}^{2}}-2\log \sin x\cdot \log \cos x=2\log \sin x\cdot \log \cos x-2\log \sin x\cdot \log \cos x\]
\[\Rightarrow {{\left( \log \cos x \right)}^{2}}+{{\left( \log \sin x \right)}^{2}}-2\log \sin x\cdot \log \cos x=0\]
Which in the form \[{{\left( a \right)}^{2}}+{{\left( b \right)}^{2}}-2a\cdot b={{\left( a-b \right)}^{2}}\]
\[\Rightarrow {{\left( \log \cos x-\log \sin x \right)}^{2}}=0\]
Since, \[\log a-\log b=\log \dfrac{a}{b}\]
\[\Rightarrow {{\left( \log \dfrac{\cos x}{\sin x} \right)}^{2}}=0\]
\[\Rightarrow \log \dfrac{\cos x}{\sin x}=0\]
\[\Rightarrow \log \cot x=0\]
Logarithm of any function is zero if that function has value is equal to 1
Hence cotx = 1
\[x={{\cot }^{-1}}1\]
\[\Rightarrow x=\dfrac{\pi }{4}\left( 4n+1 \right)\text{, n is integer}\text{.}\]
Hence, the least possible value of x is \[\dfrac{\pi }{4}\].

Note: In this problem students should not assume that we assume p and q to be positive numbers because the logarithm of any number cannot be negative. Hence, if we consider negative integers then it may be possible that we consider one positive and one negative integer to form rational but then the whole number becomes negative which is not possible. In this second problem, we should know the properties of logarithm function, trigonometric function, and inverse trigonometric function.