Question #b3dab

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Answer 1 · 2017-11-14T01:37:19

$x$ is either $1.729$ or $-2.729$ .

Use the rule of natural logarithms: $ln(a)+ln(b)=ln(a*b)$ .

$ln(x-1) + ln(x+2) = 1$
$ln((x-1)*(x+2))=1$
$ln(x^2+x-2)=1$

We all know that $ln_x x=1$ . Since the base here is $e$ , and the equation is $ln_e (x^2+x-2)=1$ , $e=x^2+x-2$ .

$x^2+x-2=e$ (Take $e$ as $2.718$ )
$x^2+x-2=2.718$
$x^2+x-2-2.718=0$
$x^2+x-4.718=0$

Here, $a=1, b=1, c=-4.718$

Use the quadratic formula: $(-b+-sqrt(b^2-4ac))/(2a)$

Solve:

$(-1+-sqrt(1^2-4*1*-4.718))/(2)$

$(-1+-sqrt(1-4*-4.718))/(2)$

$(-1+-sqrt(1+18.872))/(2)$

$(-1+sqrt(19.872))/(2),(-1-sqrt(19.872))/(2)$

$(-1+4.458)/(2),(-1-4.458)/(2)$

$(3.458)/(2),(-5.458)/(2)$

$1.729,-2.729( 3 s.f.)$

So, $x$ is either $1.729$ or $-2.729$ .