Solve $2+5\ln{x}=21$: Find $x$

[karush]

$\tiny{6.1.07 kilana HS}$
Solve $2+5\ln{x}=21$
\$\begin{array}{rlll}
\textsf{isolate} &\ln{x} &=\dfrac{19}{5} &(1)\\
\textsf{then} & &= &(2)\\
\textsf{then} & &= &(3)\\
\textsf{hence} & &= &(4)
\end{array}$
ok for (2) I presume e thru then calculate for x
just strange to see a fraction as an e exponent
tryin array on these no sure if its better :unsure:

 

[topsquark]

[math]ln(x) = a \implies x = e^a[/math]

Does it make more sense if you think of it that way?

-Dan

 

[HOI]

You understand that an exponent is just a number, don't you? An exponent can be an integer, or a fraction, an irrational number, or a complex number.

 

[karush]

roger

 

[topsquark]

You understand that an exponent is just a number, don't you? An exponent can be an integer, or a fraction, an irrational number, or a complex number.

Or even a matrix... Buwahahahaha!

-Dan

 

[karush]

https://dl.orangedox.com/QS7cBvdKw55RQUbliE

this is preliminary test to see if this works
but anyway I am trying to make a collection of 25 or more pre calc problems that came form students here on Oahu
I run a counter on the views and downloads, an earlier count it was over 50,000 on the MHB views
so it has been an awsome help. Hopefully more student from the island will join the forum when school starts up again
Im using OrangeDox because it keeps track of what happens with you posted files