What will the value of following log expressionlogxxy+logyyxI tried:logxx−logxy+logyy−logyx=1−logxy+1−logyx=2−logxy−logyxNow what after this ?

If logab=r, this means that ar=b so b=eln⁡(ar)=erln⁡(a) Therefore, ln(b) = rln (a), or logab=ln⁡(b)ln⁡a.Thus, forlogxy=ln⁡yln⁡xandlogyx=ln⁡xln⁡y,sologxy=1logyx.So:logxxy+logyyx=1−logxy+1−logyx=2−logxy−1logxy=2−(logxy+1logxy)=2−((logxy)2+1logxy)=−(logxy)2−2logx(y)+1logxy=−(logxy−1)2logxySo for r=logxy , you get −(r−1)2r. The value will depend on r; if, for example, y=x , then you get 0; if y=x2, then you get −12; if y=x−1, then you get 4, etc.

Explained: "What will the value of following log expressionlogxxy+logyyxI..."

Ben Shaffer

Answered question

2022-03-13

What will the value of following log expression

\log_{x} \frac{x}{y} + \log_{y} \frac{y}{x}

I tried:

\log_{x} x - \log_{x} y + \log_{y} y - \log_{y} x = 1 - \log_{x} y + 1 - \log_{y} x = 2 - \log_{x} y - \log_{y} x

Now what after this ?

Answer & Explanation

yourboytoyo1e

Beginner2022-03-14Added 4 answers

\log_{a} b = r

, this means that

a^{r} = b

b = e^{\ln (a^{r})} = e^{r \ln (a)}

Therefore, \ln(b) = r\ln (a), or

\log_{a} b = \frac{\ln (b)}{\ln a} .

Thus, for

\log_{x} y = \frac{\ln y}{\ln x} and \log_{y} x = \frac{\ln x}{\ln y},

\log_{x} y = \frac{1}{\log_{y} x} .

So:

\log_{x} \frac{x}{y} + \log_{y} \frac{y}{x} = 1 - \log_{x} y + 1 - \log_{y} x

= 2 - \log_{x} y - \frac{1}{\log_{x} y}

= 2 - (\log_{x} y + \frac{1}{\log_{x} y})

= 2 - (\frac{{(\log_{x} y)}^{2} + 1}{\log_{x} y})

= - \frac{{(\log_{x} y)}^{2} - 2 \log_{x} (y) + 1}{\log_{x} y}

= - \frac{{(\log_{x} y - 1)}^{2}}{\log_{x} y}

So for

r = \log_{x} y

, you get

- \frac{{(r - 1)}^{2}}{r}

. The value will depend on

r

; if, for example,

y = x

, then you get

0

; if

y = x^{2}

, then you get

- \frac{1}{2}

; if

y = x^{- 1}

, then you get 4, etc.

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