Question

solve: 7e^(2x)+2.5=20

Answer 1

0.4581

Answer 2

how did you do that?!

Answer 3

can you do something with the 2.5?

Answer 4

can't you subtract it from 20?

Answer 5

can't? why not?

Answer 6

ill take that as a yes?

Answer 7

yes u can and then divide by 7 and then take ln from both sides

Answer 8

but after that I ge stuck

Answer 9

so can we say that \(\bf 7e^{2x}+2.5=20 \implies 7e^{2x}=17.5\ \ ?\)

Answer 10

yes, then you divide both sides by 7

Answer 11

yes

Answer 12

so can we say that \(\bf 7e^{2x}+2.5=20 \implies 7e^{2x}=17.5 \implies 7e^{2x}= \cfrac{17.5}{7}\) ?

Answer 13

woops, darn Imeant \(\bf 7e^{2x}+2.5=20 \implies 7e^{2x}=17.5 \implies e^{2x}= \cfrac{17.5}{7}\) ?

Answer 14

\[e^(2x) = 2.5\]

Answer 15

and then I get stuck again

Answer 16

well, ok so \(\bf 7e^{2x}= 2.5\)
now let us use the log cancellation rule of => \(\bf log_ab^x= x\)
keeping in mind that \(\bf log_ee^x = x \)
and that \(\bf log_e = ln\)

Answer 17

well, the rule is really \(\bf \Large log_aa^x= x\)

the log base and the "result value" must be equal

Answer 18

so do you have to plug 2.5 into that somehow?

Answer 19

hmm, actually .. they don't have to be... :( anyhow, that's the cancellation rule

Answer 20

well, we could have used \(\bf log_{10}\) effectively just the same anyhow, I just happen to use ln in this case

well what you do is take log to both sides, either \(\bf log_{10}\) or \(\bf ln\) will do :)

Answer 21

so that would be \[\log_{10} e^(2x) = \log_{10}2.5\]

Answer 22

\(\bf 7e^{2x}+2.5=20 \implies 7e^{2x}=17.5 \implies e^{2x}= 2.5\\
e^{2x}= 2.5 \implies log_{10}(e^{2x}) = log_{10}(2.5)\\
\color{blue}{log_ab^x= x \implies log_{10}(e^{2x}) = 2x}\\
log_{10}(e^{2x}) = log_{10}(2.5) \implies 2x = log_{10}(2.5)\)

Answer 23

hmmm, lemme recheck myself..

Answer 24

hmmm, no I think it has to be the same base as the value, lemme rewrite that using ln :(

Answer 25

ok

Answer 26

\(\bf 7e^{2x}+2.5=20 \implies 7e^{2x}=17.5 \implies e^{2x}= 2.5\\
e^{2x}= 2.5 \implies log_{e}(e^{2x}) = log_{e}(2.5)\\
\color{blue}{log_ab^x= x \implies log_{e}(e^{2x}) = 2x}\\
log_{e}(e^{2x}) = log_{e}(2.5) \implies 2x = log_{e}(2.5)\)

Answer 27

\(\bf \color{blue}{log_aa^x= x \implies log_{e}(e^{2x}) = 2x}\)

Answer 28

so there,
2x = ln(2.5) => x = ln(2.5)/2

Answer 29

that makes a lot more sense :)

Answer 30

yw