Chena804
Solve. Please show steps.
lnxln(x1)=1



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lalaly
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use \[\ln(\frac{a}{b})=lnalnb\]

lgbasallote
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then use \(\large \log_a b = c \rightarrow a^c = b\)

lalaly
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easier way take e of both sides since its ln

lgbasallote
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as the queen orders...
\(\ln a = x \rightarrow e^x = a\)

lalaly
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hehe try it @Chena804

miat
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integration?

thunderemperor
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\[\ln x\ln (x1)=1\]
\[\ln (x \div(x1))=1\]
\[\left(\begin{matrix}x\\ x1\end{matrix}\right)=e ^{1}\]
\[x=(x1)e ^{1}\]
You can solve the rest. ;)