e^ln x = 3 QUICK HELP??? I don't remember how to do this but I may think the answer is e^3
have you covered logarithms yet?
Yeah, but I am a little rusty since it was quite a while ago.
ok so... let's us use the log cancellation rule of \(\Large { \bf {\color{brown}{ a}}^{log_{\color{brown}{ a}}x}=x\qquad thus \\ \quad \\ e^{ln(x)}\implies {\color{brown}{ e}}^{log_{\color{brown}{ e}}(x)}=3\implies ? }\)
3^3? Which equals 27
?
I'm not sure. haha
well..... look at the log cancellation rule above :)
can the left-side by simplified?
Alright I'll look it over
OH, is it log3? because of the e's cancel??
hmm ahemm... \(\Large \bf \textit{log cancellation rule }\to {\color{brown}{ a}}^{log_{\color{brown}{ a}}(whatever)}=whatever\)
if the base of the log, MATCHES the base of the expression.... then... :)
3^log3? Sorry, I feel really dumb >.<
notice the "a", they MATCH don't they? and if they do... notice what's the equation in the cancellation rule
Yeah, the a's match. since it is a^loga3???? = 3?
welll if you had a 3 there.. .yes
Okay, that makes sense I: I think there was an x?
\(\textit{log cancellation rule }\to \Large {{\color{brown}{ a}}^{log_{\color{brown}{ a}}(x)}=x\qquad thus \\ \quad \\ e^{ln({\color{blue}{ x}})}\implies {\color{brown}{ e}}^{log_{\color{brown}{ e}}({\color{blue}{ x}})}=3\implies {\color{blue}{ x}}=3 }\)
Yeah thats what I had up there^^^ xD Thank you soooooo much you helped a lot :)
yw
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