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Mathematics 20 Online
OpenStudy (chaotic_butterflies):

ln(3x+7)=1.31 (ln as in log natural)

OpenStudy (retireed):

First raise both sides by e and solve for x... e^[ln(3x+7)] = e^1.31 3x + 7 = e^1.31

OpenStudy (chaotic_butterflies):

I was afraid that might happen.. alright let me try to write it out

OpenStudy (chaotic_butterflies):

(and thank you for helping)

OpenStudy (chaotic_butterflies):

I got \[x\approx -1.6389\]

zepdrix (zepdrix):

Ya? Hmm my value is coming out slightly different..

zepdrix (zepdrix):

Subtract 7, divide 3,\[\large\rm x=\frac{e^{1.31}-7}{3}\]This is what you input? Something similar?

OpenStudy (retireed):

i got closer to-1.097

OpenStudy (chaotic_butterflies):

I simplified e^1.31 to 2.0834 and solved algebraically from there

OpenStudy (chaotic_butterflies):

Honestly my teacher isn't a stickler, I don't think I'm too far off from your answer @retirEEd

OpenStudy (chaotic_butterflies):

Oh wait a sec, I calculated e^1.13

OpenStudy (chaotic_butterflies):

Whoops

zepdrix (zepdrix):

Ah :D

OpenStudy (chaotic_butterflies):

I got -1.0979

OpenStudy (chaotic_butterflies):

So I got what you did @retirEEd

OpenStudy (retireed):

Try to plug in and solve for @zebdrix equation. Personally, I think your answer is significantly different. I solved mine by graphing the original equation and seeing where it equalled 1.31.

OpenStudy (retireed):

Sorry I'm watching TV so my responses are running behind your answers.

OpenStudy (chaotic_butterflies):

That's fine, I'm a bit scatterbrained myself. Thanks for the help @retirEEd @zepdrix

OpenStudy (retireed):

np

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