solve the given equation. round to the nearest hundreth, if neccesary 9e^x-9=10??
@ganeshie8 ??
\(9e^x-9=10\)
like that ?
yes
start by adding \(9\) both sides
\(\large \bf \textit{log cancellation rule of }log_{\color{red}{ a}}({\color{red}{ a}}^x)=x \\ \quad \\ 9e^x-9=10\implies 9e^x=19\implies e^x=\cfrac{19}{9} \\ \quad \\ \large log_{\color{red}{ e}}({\color{red}{ e}}^x)=log_{\color{red}{ e}}\left(\cfrac{19}{9}\right)\) what do you think?
so the answer would be 2.11??
hmm 2.11?
use the log cancellation rule on the left side
so what would be the answer @jdoe0001
dunno what would the left side give you using the cancellation rule?
x=2.11
well, you're correct, the left side ends up as "x" but on the right side, you're skipping the \(\bf ln \) part
so what would ii do??
@jdoe0001
\(\large \bf log_{\color{red}{ e}}({\color{red}{ e}}^x)=log_{\color{red}{ e}}\left(\cfrac{19}{9}\right)\implies x=log_{\color{red}{ e}}\left(\cfrac{19}{9}\right)\implies x=ln\left(\cfrac{19}{9}\right)\)
what do you get as the answer??
@jdoe0001
hmmm do you have an [ln] button in your calculator? you should I'd think
i dont have a caculator @jdoe0001
natural logarithm, or "ln" should be there under "functions > calc"
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