Question

if xy - x^x then 1/(logx y) + 1) =​

Answer 1

Answer:

1/x

Step-by-step explanation:

xy=x^x

y=(x^x)/x

Logging both sides

log(y)= log((x^x)/x)

log(y)= logx^x - logx

= xlogx - logx

= x×1 - 1

= x - 1

Using log(y) we have,

1/(log(y)+1) = 1/(x-1+1)

= 1/x

Answer 2

Note - Question should be like this, if $\[xy = {x^x}\]$ then $\[\frac{1}{{\log xy}} + 1 = \]$​

Given,

$\[xy = {x^x}\]$--------(1)

$\[\frac{1}{{\log xy}} + 1\]$------(2)

Solution,

Know that,

$\[xy = {x^x} - - - \left( 1 \right)\]$

Taking log on both side,

$\[\log xy = \log {x^x}\]$

$\[ \Rightarrow \log xy = x\log x\]$

Put this value in equation 2,

$\[ = \frac{1}{{\log xy}} + 1\]$

$\[ = \frac{1}{{x\log x}} + 1\]$

$\[ = \frac{{1 + x\log x}}{{x\log x}}\]$

Hence the value is $\[\frac{{1 + x\log x}}{{x\log x}}\]$.