Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

solve the given equation. round to the nearest hundreth, if neccesary 9e^x-9=10??

OpenStudy (anonymous):

@ganeshie8 ??

ganeshie8 (ganeshie8):

\(9e^x-9=10\)

ganeshie8 (ganeshie8):

like that ?

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

start by adding \(9\) both sides

OpenStudy (jdoe0001):

\(\large \bf \textit{log cancellation rule of }log_{\color{red}{ a}}({\color{red}{ a}}^x)=x \\ \quad \\ 9e^x-9=10\implies 9e^x=19\implies e^x=\cfrac{19}{9} \\ \quad \\ \large log_{\color{red}{ e}}({\color{red}{ e}}^x)=log_{\color{red}{ e}}\left(\cfrac{19}{9}\right)\) what do you think?

OpenStudy (anonymous):

so the answer would be 2.11??

OpenStudy (jdoe0001):

hmm 2.11?

OpenStudy (jdoe0001):

use the log cancellation rule on the left side

OpenStudy (anonymous):

so what would be the answer @jdoe0001

OpenStudy (jdoe0001):

dunno what would the left side give you using the cancellation rule?

OpenStudy (anonymous):

x=2.11

OpenStudy (jdoe0001):

well, you're correct, the left side ends up as "x" but on the right side, you're skipping the \(\bf ln \) part

OpenStudy (anonymous):

so what would ii do??

OpenStudy (anonymous):

@jdoe0001

OpenStudy (jdoe0001):

\(\large \bf log_{\color{red}{ e}}({\color{red}{ e}}^x)=log_{\color{red}{ e}}\left(\cfrac{19}{9}\right)\implies x=log_{\color{red}{ e}}\left(\cfrac{19}{9}\right)\implies x=ln\left(\cfrac{19}{9}\right)\)

OpenStudy (anonymous):

what do you get as the answer??

OpenStudy (anonymous):

@jdoe0001

OpenStudy (jdoe0001):

hmmm do you have an [ln] button in your calculator? you should I'd think

OpenStudy (anonymous):

i dont have a caculator @jdoe0001

OpenStudy (jdoe0001):

-> http://desmos.com/calculator <----

OpenStudy (jdoe0001):

natural logarithm, or "ln" should be there under "functions > calc"

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!