Question

Solve for x :
log (x+1)+ log (x-1)=1
​

Answer 1

it is in the form of Loga+logb

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

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

$log {x}^{2} - log {1}^{2 } = 1$

$2logx - 2log1 = 1$

2logx -0=1

logx=1/2

x=log inverse of 1/2