Question

Show that, $\log[\sqrt{x^{2}+1}+x] + \log[\sqrt{x^{2}+1}-x]=0$

Answer 1

Answer:

Step-by-step explanation:

Hi,

To Prove ㏒[√(x² + 1) + x] + ㏒[√(x² + 1) - x] = 0

Consider L.H.S = ㏒[√(x² + 1) + x] + ㏒[√(x² + 1) - x]

We know from Additive Property of logarithms,

㏒ m + ㏒ n = ㏒ (mn)

So, ㏒[√(x² + 1) + x] + ㏒[√(x² + 1) - x]

= log[(√(x² + 1) + x)*(√(x² + 1) - x)]

= ㏒[ (√(x² + 1) )² - x²] = 0

= ㏒ [ x² + 1 - x² ]

= ㏒ 1

But ㏒ 1 = 0,

Thus, L.H.S = 0

= R.H.S

Hence, Proved !

Hope, it helps !

Answer 2

Solution :

***************************************
We know that ,

i ) $log_{a}x + log_{a}y = log_{a}xy$

ii ) $log_{a}a=0$

*****************************************

LHS = $\log[\sqrt{x^{2}+1}+x] + \log[\sqrt{x^{2}+1}-x]$

=$\log[\sqrt{x^{2}+1}+x] \times [\sqrt{x^{2}+1}-x]$

= $\log[\sqrt{x^{2}+1}^{2}-x^{2}]$
= $log[(x^{2}+1)-x^{2}$

= $log( x^{2} + 1 - x^{2} )$

= log 1

= 0

= RHS

•••••