Question

How do you solve (8e^(8x)) + (-e^(-x))=0 ?? How do you solve (8e^(8x)) + (-e^(-x))=0 ?? @Mathematics

Answer 1

\[8e^{8x}-e^{-x}=e^{-x}(8e^{9x}-1)=0\]

Answer 2

set both factors =0
that first factor will never be 0
bt the second factor will be 0

Answer 3

\[8e^{9x}-1=0 => 8e^{9x}=1 => e^{9x}=\frac{1}{8}\]

Answer 4

\[\ln(e^{9x})=\ln(\frac{1}{8})=> 9x=\ln(\frac{1}{8})=x=\frac{1}{9} \ln(\frac{1}{8})\]

Answer 5

why do you get \[8^{9x} = e ^{-x} \] when you factor out?

Answer 6

(not =, sorry)

Answer 7

what?

Answer 8

when you factor out e^-x, how do you get 8e^9x ?

Answer 9

remember your law of exponents
\[e^{8x}=e^{9x-1x}=e^{9x}e^{-x}\]

Answer 10

\[8e^{8x}-e^{-x}=8e^{9x-x}-e^{-x}=8e^{9x}e^{-x}-e^{-x}=e^{-x}(8e^{9x}-1)\]

Answer 11

8=9-1

Answer 12

oh okay! thank you

Answer 13

you gots it for reals?

Answer 14

yeah, i always forget that you can split them apart

Answer 15

if you don't mind me asking, how would you solve -10(cosx)(sinx) - 10(cosx) = 0 ? since you're good at them

Answer 16

i know that you can first factor out (-10cosx)

Answer 17

\[-10\cos(x)(\sin(x)+1)= 0=> \cos(x)=0 \text{ or} \sin(x)+1=0 \]

Answer 18

i see. thank you!