Question

if \(\Large x\geq y\) and \(\Large y>1\) , then value of the expression


\(\Large\tt \begin{align} \color{black}{\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}
}\end{align}\)

can never be
\(a. -1\\
b. 0.5\\
c. 0\\
d. 1\)

Answer 1

i got this
\(\Large\tt \begin{align} \color{black}{\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x} \\
=\log_x (x)- \log_x (y)+\log_y (y)+\log_y (x)\\
=2- \log_x (y)-\log_y (x)\\
=2- \dfrac{\log_y (y)}{\log_y (x)}-\log_y (x)\\
=2- \dfrac{1}{\log_y (x)}-\log_y (x)\\
=2- \dfrac{1}{\log_y (x)}-\dfrac{(\log_y (x))^2}{\log_y (x)}\\
=2- \dfrac{1}{\log_y (x)}-\dfrac{2\log_y (x)}{\log_y (x)}\\
=-\dfrac{1}{\log_y (x)}
}\end{align}\)

Answer 2

@campbell_st

Answer 3

is my method wrong

Answer 4

After some steps, I found the equivalent form of your expression, here is:
your expression = \[-\frac{ (\log _{x}y-1)^{2} }{ \log _{x}y}\], from which, I excluded the values 0.5 and 1

Answer 5

that's right?

Answer 6

yes thats right but u swapped x and y

Answer 7

I have changed the base of the logarithm at the second term, from x to y!

Answer 8

ok but i dont get how u excluded options 0.5 and 1

Answer 9

In my expression, please change variable in this way:
\[\log _{x}y=k\]
being k the new variable, now write my expression as a function of k, namely:
\[-\frac{ (k-1)^{2} }{ k }\]
finally, substitute to my expression your values, you will find an quadratic equation in k, and in some cases there is no solutiopn for k.

Answer 10

Sorry I wanted say, equate my expression to your values...

Answer 11

m still confused

Answer 12

the 3rd lines copied wrong

Answer 13

it's 2-log_x (y)+ log_y (x) , you did - for the last term

Answer 14

for example, for value -1, we can write:
\[-\frac{ (k-1)^{2} }{ k }=-1\]
now try to solve that equation for k, please

Answer 15

however, I don't know why you have to do a long argument like that,

Answer 16

log > 0 for all x, y
the first one is >0, the second one is >0 also, how the sum <0 (I mean option a)

Answer 17

get me? positive number + positive number CANNOT BE negative one
then, the so-called "Never be" is a)

Answer 18

no?? hihihi....

Answer 19

is this correct
\(\Large\tt \begin{align} \color{black}{\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x} \\
=\log_x (x)- \log_x (y)+\log_y (y)-\log_y (x)\\
=2- \log_x (y)-\log_y (x)\\
=2- \dfrac{\log_y (y)}{\log_y (x)}+\log_y (x)\\
=2- \dfrac{1}{\log_y (x)}+\log_y (x)\\
=2- \dfrac{1}{\log_y (x)}+\dfrac{(\log_y (x))^2}{\log_y (x)}\\
=2- \dfrac{1}{\log_y (x)}+2\dfrac{\log_y (x)}{\log_y (x)}\\
=-\dfrac{1}{\log_y (x)}
}\end{align}\)

Answer 20

nope

Answer 21

why :|

Answer 22

i think the second line was wrong third was correct

Answer 23

btw what is min and max value for any log

Answer 24

\(log _x y^2 = 2log_x y\)
but \((log_x y)^2 \neq 2log_x y\)
for example
\(log_{10} 100 =2 \), that is \(\log_{10}10^2=2log_{10}10\)

Answer 25

but \((log_{10} 10)^2 =1\)

Answer 26

so that, you can't take down 2 on the 6th line

Answer 27

@Michele_Laino \(k ~will ~be+~\)\)

Answer 28

the values of k are not important, what is important is if the equation admits or no, solutions for k

Answer 29

\[\large \rm \log (x^2) = 2\log(x) \ne \left[\log(x)\right]^2\]

Answer 30

ok i got it :)

Answer 31

\[\large \rm \left[\log(x)\right]^2 = \log(x) \times \log(x)\]

Answer 32

\[\large \rm \log(x^2) = \log(x\cdot x) = \log(x) + \log(x)\]