How do you solve $ln \sqrt{x + 2} = 1$ $lnsqrt(x+2)=1$ ?

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How do you solve $ln \sqrt{x + 2} = 1$ $lnsqrt(x+2)=1$ ?

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Answer 1 · 2018-06-16T12:03:13

THe solution is $= e^{2} - 2$ $=e^2-2$

Explanation:

Solve the equation as follows :

$ln \sqrt{x + 2} = 1$ $lnsqrt(x+2)=1$

$\Rightarrow$ $=>$ , ${ln (x + 2)}^{\frac{1}{2}} = 1$ $ln(x+2)^(1/2)=1$

$\Rightarrow$ $=>$ , $\frac{1}{2} ln (x + 2) = 1$ $1/2ln(x+2)=1$

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

$\Rightarrow$ $=>$ , $x + 2 = e^{2}$ $x+2=e^2$

$\Rightarrow$ $=>$ , $x = e^{2} - 2 = 5.389$ $x=e^2-2=5.389$

Answer 2 · 2018-06-16T12:13:52

$x = e^{2} - 2 \approx 5.3890560989$ $x=e^2-2~~5.3890560989$

Explanation:

Remember that $ln (a) = c$ $ln(a)=c$ means ${log}_{e} (a) = c$ $log_e(a)=c$
$XXXXXXXXXXXXXXXXXX \Rightarrow e^{c} = a$ $color(white)("XXXXXXXXXXXXXXXXXX") rArre^c=a$

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

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

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

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How do you solve $ln \sqrt{x + 2} = 1$ $lnsqrt(x+2)=1$ ?

2 Answers

Explanation:

Explanation:

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