How do you solve $log ( x +1 ) + log 7 = log 14 - log ( 2-x )$ ?

Question

3946 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-12T13:15:36

$x=1$

$log(x+1)+log7=log14-log(2-x)$

$log(x+1)+log(2-x)=log14-log7$

$log((x+1)/(2-x))=log2$

$(x+1)/(2-x)=2$

$x+1=4-2x$

$3x=3$

$x=1$

Recall:

$log_10 a + log_10 b = log_10(atimesb) = log_10 ab$

$log_10 a- log_10 b=log_10(adivide b)=log_10 (a/b)$

How do you solve log ( x +1 ) + log 7 = log 14 - log ( 2-x ) log ( x +1 ) + log 7 = log 14 - log ( 2-x ) ?