Solving Natural Logs

[Angela123]

Solve ln (x-1)^(1/2)=2

=e* ln (x-1)^(1/2)=2*e

=((x-1)^(1/2)^2=2e^2

=(x-1)=2e^2

x=(2e^2)+1

=6.44

Is this right?

 

[Deleted member 4993]

Solve ln (x-1)^(1/2)=2

=e* ln (x-1)^(1/2)=2*e<<< How's that --> the right hand side should be e[sup:1oqefd7f]2[/sup:1oqefd7f]

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ln[(x-1)][sup:1oqefd7f]1/2[/sup:1oqefd7f] = 2

1/2 * ln(x-1) = 2 <<< Law of logarithm

ln(x-1) = 4

x - 1 = e[sup:1oqefd7f]4[/sup:1oqefd7f]

x = 1 + e[sup:1oqefd7f]4[/sup:1oqefd7f]

___________________________________________


=((x-1)^(1/2)^2=2e^2

=(x-1)=2e^2

x=(2e^2)+1

=6.44

Is this right?

 

[pka]

\(\displaystyle \begin{gathered} \ln (x - 1)^{1/2} = \left( {1/2} \right)\ln (x - 1) = 2\; \Rightarrow \;\ln (x - 1) = 4 \hfill \\ x - 1 = e^4 \; \Rightarrow \;x = e^4 + 1 \hfill \\ \end{gathered}\)

 

[BigGlenntheHeavy]

$\displaystyle ln(x-1)^{1/2} \ = \ 2$

$\displaystyle (x-1)^{1/2} \ = \ e^{2}$

$\displaystyle x-1 \ = \ e^{4 }$

$\displaystyle x \ = \ e^{4} +1 \ = \ 55.5981500331$

$\displaystyle Check: \ ln(e^{4}+1-1)^{1/2} \ = \ 2$

$\displaystyle ln(e^{4})^{1/2} \ = \ 2$

$\displaystyle ln(e^{2}) \ = \ 2, \ 2 \ = \ 2$