Precalculus?

Question

how would you solve...

1.2e^(-5x)+2.6=3

Answer 1

1.2e^(-5x) + 2.6 = 3
1.2e^(-5x) = .4
e^(-5x) = .4/(1.2)
e^(-5x) = (1/3)
-5x = ln(1/3)
x = (ln(1/3))/(-5)
x = -(ln(1/3))/(-5)

ANS : (ln(3))/5 or about .21972

Answer 2

Let me see if I can do this...

first subtract 2.6...

1.2e^(-5x)=0.4

then divide 1.2...

e^(-5x)=0.33

then take the natural log of both sides...

ln(e^(-5x))=ln(0.33)

from logrithmic properties, the ln on the left side would cancel out...

-5x=ln(0.33)

divide by -5...

and x=0.2217...

I hope I did that correctly. Check with someone else first.

*haha, I guess someone else got to it before I did.

Answer 3

Move the 2.6 to the other side of the equal sign.
2e^(-5x) = 0.4

Take the natural log of both sides.
ln (2e^(-5x)) = ln 0.4

Simplify the left side.
ln 2 + ln e^(-5x) = ln 0.4
ln e^(-5x) = ln 0.4 - ln 2

The natural log of e^(-5x) is -5x, because ln is log base e.
-5x = ln (0.4 / 2)
-5x = ln (0.2)
x = (1/-5) ln (0.2)

Sorry, I don't have a calculator that can do natural logs with me at the moment.

Answer 4

1.2^5x) = 1.2+2.6=3.8-5x=3 x=8x5=40 1.2^=37 40-37=3

Answer 5

1.2e^(-5x)+2.6=3
take ln the both sides.
lne^-6x=ln(0.4)
we know that:
lne^a =a
-6x=-0.91 or x=-0.91/-6
x=0.219922558

Answer 6

Solve the (-5x).

Answer 7

All the numbers! Just erase them. Just get rid of them. The variables, the decimals, EVERYTHING! They kill my eyes. Die numbers, die!

Answer 8

maybe it true wat they say "precalculus is hard than calculus" i'm takin calculus honors and i have no idea wat this means lol, maybe because precal breaks everything down step by step