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Mathematics 23 Online
OpenStudy (anonymous):

in(1-3x)+3=9

ganeshie8 (ganeshie8):

(1-3x)+3=9 its like this ?

OpenStudy (anonymous):

yes and you have to use that log thing natural log

ganeshie8 (ganeshie8):

oh ok.. \(ln(1-3x) + 3 = 9\) like this right ?

OpenStudy (anonymous):

yes :D correct can you explain how to do it as well

ganeshie8 (ganeshie8):

\(\ln(1-3x) + 3 = 9\) \(\ln(1-3x) = 6\)

ganeshie8 (ganeshie8):

we must know below thing, to proceed further : \(\huge \ln x = a \\ \huge=> x = e^a\)

ganeshie8 (ganeshie8):

you familiar with above formula ?

OpenStudy (anonymous):

no

ganeshie8 (ganeshie8):

Ok.. can you pls tell what you know about logs ?

ganeshie8 (ganeshie8):

so that we can see how to go about the solution.. but there is only one way to solve this problem using above mentioned formula :\

OpenStudy (anonymous):

i was using a plug in formula for it but the answers i got didnt match and not much :/ i watched a few youtube tutorials on them but nothing helped me further my understanding towards this

ganeshie8 (ganeshie8):

Okay. lets use the formula and move ahead. il explain the formula in the end ok

OpenStudy (anonymous):

okay than thank you very much appreciate this

ganeshie8 (ganeshie8):

\(\ln(1-3x) = 6\) \(1-3x = e^6\) \(3x = 1-e^6\) \(x = \frac{1-e^6}{3}\)

ganeshie8 (ganeshie8):

is that the answer or ur textbook converted it into numberical

OpenStudy (anonymous):

i think it is converted

ganeshie8 (ganeshie8):

whats the answer ?

OpenStudy (anonymous):

yes it was :)

OpenStudy (anonymous):

but im a little lost on the last part can you explain it

ganeshie8 (ganeshie8):

great :)

ganeshie8 (ganeshie8):

ok

ganeshie8 (ganeshie8):

what point u stuck at

ganeshie8 (ganeshie8):

\(\ln(1-3x) = 6\) \(1-3x = e^6\) here ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

sorry my computers actin slow i have 2 keep refreshing it

ganeshie8 (ganeshie8):

np :)

OpenStudy (anonymous):

thank u for understanding and helping

ganeshie8 (ganeshie8):

\(\ln(1-3x) = 6\) can also be written as : \(\log_e(1-3x) = 6\) now we apply the formula : \(log_a x = c => x = a^c\)

OpenStudy (anonymous):

mmhmm

ganeshie8 (ganeshie8):

i understand u... but im failing to explain eloquently lol

OpenStudy (anonymous):

its fine i get it somewhat ill be back with another qustion though

ganeshie8 (ganeshie8):

Ok.. may be someone else can pitch in and help u understand @phi

ganeshie8 (ganeshie8):

@mahmit2012

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