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

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Answer 1 · 2015-11-18T17:19:07

$x = - 4 or + 3$ $x=-4 " or " +3$

Let "log base" be b

Known that ${log}_{b} (b) = 1$ $log_b(b)=1$

Assumption: The use of 'log' in the question is referring to logs to base 10 giving:

$log_10(x+2) +log_10(x-1) =1.....................(1)$

But $color(white)(xxxxxx)log_10(10)=1 .............................(2)$

Substitute (2) into (1) giving:

$log_10(x+2)+log_10(x-1) = log_10(10) ..........(3)$

But $log_10(x+2)+log_10(x-1) = log_10[ (x+2)(x-1)]$

Rewrite (3) as:

$log_10[(x+2)(x-1)] = log_10(10)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Personal note: I am of 'very' old school where $"antilog" -> log^(-1)$

Taking antilogs $-> (x+2)(x-1)=10$

$=> x^2+x-12=0$

$(x +4)(x-3)=0$

$color(blue)( x=-4 " or " +3)$

How do you solve Log(x+2)+log(x-1)=1log(x+2)+log(x−1)=1Log(x+2)+log(x-1)=1?