The value of the logarithmic expression can never be $\ldots$

Question

The value of the logarithmic expression $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x},\text{where}\quad x\geq y>1\quad$ can never be

$\bf\text{options}$ a.) $-1\quad$ b.)$\quad0.5\quad$ c.) $\quad0\quad$ d.) $\quad1$

$$\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\\ &=\frac{2\log_x y}{\log_x y}-\frac{(\log_x y)^2}{\log_x y} -\frac{1}{\log_x y}\\ &=-\frac{ (\log _{x}y-1)^{2} }{ \log _{x}y}\end{align}$$

here if i input $x=y=2$ then option $0$ is removed , not sure how to deduce the rest.

Wolfram does gives the answer as option $1$

Answer 1

$$\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\\ &=2-\dfrac{\ln y}{\ln x} -\dfrac{\ln x}{\ln y}\\ &=2-\dfrac{\ln² y+\ln²x}{\ln x \ln y} \\ &=-\dfrac{(\ln y-\ln x)²}{\ln x \ln y}\end{align}$$

this value does never equal ${1,0.5}$

Answer 2

In light of @John's leading comment, if we put $$A=\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}$$ then $$A=2-\left(\log_x y +\log_y x\right)=2-(z+\frac{1}{z})\le 0,~~~z=\log_x y$$ This means that another option could be chosen. In fact, using maple says that: it can never be $0.5$ also:

 > implicitplot(log[x](x/y)+log[y](y/x) = 1/2, x = 1 .. 100, y = 1 .. 100);

Answer 3

Note: We start in the same way as OP did and then we proceed accordingly.

We consider

\begin{align*} \log_x\frac{x}{y}+\log_y\frac{y}{x}\qquad\qquad\qquad x\geq y>1\tag{1} \end{align*}

• Validity of domain: At first, we check the domain and observe that since $\log_bz$ is defined for any $z>0$ and for $b>0, b\ne 1$ the expression (1) is valid for all $x\geq y>1$.

• Since $\log_bx=\frac{\log_kx}{\log_kb}$ for any bases $b,k>1$ we obtain \begin{align*} \log_xy=\frac{\log_yy}{\log_yx}=\frac{1}{\log_yx}\tag{2} \end{align*}

• We also keep in mind that for $x\geq y > 1$ \begin{align*} 0<\log_xy\leq 1 \tag{3} \end{align*}

Options b.) and d.)

We start now similarly as OP did: \begin{align*} \log_x&\frac{x}{y}+\log_y\frac{y}{x}\\ &=\log_xx-\log_xy+\log_yy-\log_yx\\ &=2-\log_xy-\frac{1}{\log_xy}\\ &=-\frac{1}{\log_xy}\left((\log_xy)^2-2\log_xy+1\right)\\ &=-\frac{\left(\log_xy-1\right)^2}{\log_xy}\leq 0\tag{4} \end{align*}

In (4) we use $0<\log_xy\leq 1$ accordingly to (3).

Conclusion: The expression (1) is less or equal to zero whenever $x\geq y >1$. Therefore it can never be option b.) with value $0.5$ or option d.) with value $1$

Options a.) and c.)

We now set $z=\log_xy$ and consider accordingly to (4) the expression

\begin{align*} -\frac{\left(z-1\right)^2}{z}=a\qquad\qquad a \in \mathbb{R} \end{align*} or equivalently \begin{align*} z^2-(a-2)z+1=0 \end{align*} with solutions \begin{align*} z_{1,2}=1-\frac{a}{2}\pm\frac{1}{2}\sqrt{a(a-4)} \end{align*} Now, if we set accordingly to option c.) $a=0$ we see $z_{1,2}=1$ and therefore $$\log_xy=1 \Rightarrow x=y$$

Conclusion: Letting $x=y>1$, we get

\begin{align*} \log_x&\frac{x}{y}+\log_y\frac{y}{x}\\ &=\log_x\frac{x}{x}+\log_x\frac{x}{x}\\ &=2\log 1\\ &=0 \end{align*} and so we can't skip option c.)

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Next we set $a=-1$ and get $z_{1,2}=\frac{1}{2}(3\pm \sqrt{5})$

Conclusion: Since $z_2=\frac{1}{2}(3-\sqrt{5})< 1$ we see that according to (3) we also find a solution for option a.) with value $-1$

Answer 4

\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\end{align}

Let $y^p = x$, therefore $\log_y x =p$ and $ \log_x y = \frac1p $ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$ So, $1-\log_x y +1-\log_y x = 2 - (p + \frac1p)$

For the RHS of above expression to equate to -1, 0.5, 0, 1, we must have p + $\frac1p$ = 3, 1.5, 2, 1 respectively. So, we have to solve for $p$ the corresponding quadratic equations which work out to be $\quad$ $p^2-3p+1=0$, $\quad$ $2p^2-3p+2=0$, $\quad$ $p^2-2p+1=0$, $\quad$ $p^2-p+1=0$. $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ On solution we obtain $p=(3+\sqrt{5})/2 $ & $p=(3-\sqrt{5})/2$ for the first, $\quad$$\quad$ $p=1$ & $p=\frac12$ for the second, $\quad$$\quad$ $p=1$ for the third, $\quad$$\quad$ and $p=(1+\sqrt3i)/2$ & $p=(1-\sqrt3i)/2$ for the fourth.

Since $y^p = x$, imaginary $y$ would be out of domain of the problem and also make $x$ outside the domain of the problem (domain is x,y > 1, i.e. x, y real) and hence solution $p=1$ which corresponds to $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}=1$ is not possible. So, answer is option d.) $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ Also notice that for second equation, $p=1$ is a false root, and $p=\frac12$ would make $y \gt x$ which is also outside the domain of the problem. So, this equation which corresponds to $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}=\frac12$ is also not possible

Answer 5

In this question, take $log_y x=a$. Then the expression$(E)$ given in the question reduces to $$E=2-(a+\frac 1a)$$ Now $a$ and $\frac1a$ are both positive or both negative. Since $x,y>1$ $a$ is positive. Using $A.M-G.M$ inequality you can say that $$a+\frac 1a \ge 2\sqrt{a.\frac 1a}\ge 2.$$Hence $$E=2-(a+\frac 1a)\le 2-2\le 0$$ Hence options $(b)$ and $(d)$ are ruled out. For $E=0$, equality must hold in the inequality, i.e., $a=1$. This is possible when $x=y$. For $(a)$ to be true $$a+\frac 1a=3\implies a^2-3a+1=0\implies a=\frac {3\pm\sqrt5}2$$But $x>y\implies a>1$. Therefore $a=\frac {3+\sqrt5}2$. This is very well possible. Hence ($a$) and ($c$) are correct.