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Mathematics 19 Online
OpenStudy (anonymous):

Solve: 5.8=6e^2x+3=8

OpenStudy (anonymous):

\[5.8 = 6*e^{2*x}+3=8\] is this the equation?

OpenStudy (bahrom7893):

wait what?

OpenStudy (anonymous):

where did I come up with that? Lol. No, the equation is actually 5.8=6e^2x+5

OpenStudy (bahrom7893):

okay now subtract 5 from both sides, what do you get?

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

0.8=6e^2x

OpenStudy (bahrom7893):

Okay now divide bot sides by 6, what do u get?

OpenStudy (anonymous):

.1333=e^2x

OpenStudy (bahrom7893):

okay now take Ln of both sides.

OpenStudy (bahrom7893):

Ln(0.133)=Ln(e^2x)

OpenStudy (bahrom7893):

Now there's this rule: Lna^b=bLna, apply it to the right side, what do u get?

OpenStudy (anonymous):

huh?

OpenStudy (anonymous):

\[ln(a^{b}) = b*ln(a)\]

OpenStudy (bahrom7893):

Ln(e^(2x))right Ln(a^(b)) = b*Ln(a) let your "a" be e and "b" be 2x

OpenStudy (bahrom7893):

So just use that rule: e is your a and 2x is your b

OpenStudy (bahrom7893):

okay look: Ln(e^2x) = lne^(2x)=(2x)Lne

OpenStudy (bahrom7893):

Ln(e)=1 Memorize this rule

OpenStudy (anonymous):

How do I enter the equation on my calculator to get the final answer?

OpenStudy (bahrom7893):

So (2x)*Ln(e), where ln(e) = 1 what does that equal to?

OpenStudy (bahrom7893):

Solve(5.8=6e^2x+5,x)

OpenStudy (anonymous):

0?

OpenStudy (bahrom7893):

That's how u enter it

OpenStudy (anonymous):

I guess I'm not following. :-( The last place I kind of understood is when I divided both sides by 6 and came up with .1333=e^2x

OpenStudy (bahrom7893):

Okay now, see the e^(2x)? the 2x is in the power, so u need to take Ln on both sides to get it down, ln meaning natural log

OpenStudy (bahrom7893):

So: Ln(0.1333)=Ln(e^(2x))

OpenStudy (bahrom7893):

now Lne^a=a that's a rule Ln(0.1333)=2x x = Ln(0.1333)/2 = whatever the calculator gives u..

OpenStudy (anonymous):

Ah ha! -1.0075

OpenStudy (anonymous):

Want to help me with another equation (the last one) or am I too big of a headache? Lol.

OpenStudy (bahrom7893):

okay sure post it as a new thread, this is getting long and slow..

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