MHB Solve $2+5\ln{x}=21$: Find $x$

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To solve the equation $2 + 5\ln{x} = 21$, isolate $\ln{x}$ to find $\ln{x} = \frac{19}{5}$. The next step involves exponentiating both sides to express $x$ in terms of $e$. The discussion highlights the understanding that exponents can take various forms, including fractions. Additionally, there is a mention of creating a collection of precalculus problems for students, aiming to engage more learners in the forum. This collaborative effort is supported by tracking views and downloads of the shared materials.
karush
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$\tiny{6.1.07 kilana HS}$
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:
 
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[math]ln(x) = a \implies x = e^a[/math]

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

-Dan
 
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.
 
roger
 
Country Boy said:
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
 
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
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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