How do you solve $lnsqrt(x+2)=1$ ?

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Answer 1 · 2018-06-16T12:03:13

THe solution is $=e^2-2$

Solve the equation as follows :

$lnsqrt(x+2)=1$

$=>$ , $ln(x+2)^(1/2)=1$

$=>$ , $1/2ln(x+2)=1$

$=>$ , $ln(x+2)=2$

$=>$ , $x+2=e^2$

$=>$ , $x=e^2-2=5.389$

Answer 2 · 2018-06-16T12:13:52

$x=e^2-2~~5.3890560989$

Remember that $ln(a)=c$ means $log_e(a)=c$
$color(white)("XXXXXXXXXXXXXXXXXX") rArre^c=a$

So
$color(white)("XXX")ln(sqrt(x+2))=1$
means
$color(white)("XXX")e^1=sqrt(x+2)$

and therefore
$color(white)("XXX")e^2=x+2$
and
$color(white)("XXX")x+2(color(magenta)(-2))=e^2color(magenta)(-2)$
or
$color(white)("XXX")x=e^2-2$

...an approximation to this value can be determined using a calculator.

How do you solve lnsqrt(x+2)=1lnsqrt(x+2)=1?