Question

solve for x...
(let me type it as an equation)

Answer 1

\[y=(10^{x}+10^{-x})\div8\]

Answer 2

we can do this

Answer 3

multiply by 8 to get
\[8y=10^x+10^{-x}\] multiply both sides by
\[10^x\] so get
\[8y\times 10^x=10^{2x}+1\] put all terms on one side get
\[10^{2x}-8y\times 10^x+1=0\]

Answer 4

then you have a quadratic equation in
\[10^x\] in other words you can thing of this as
\[z^2-8yz+1=0\] and you can solve it using the quadratic formula with
\[a=1,b=-8y, c = 1\]

Answer 5

you get
\[z=\frac{8y\pm\sqrt{64y^2-4}}{2}\]

Answer 6

i got that and then simplifited to...

x= 4y + (12y^2-1)^)(1/2)
(only positive answer exists because of the rules of logs)

i'm doing this question online, and when i input this answer, it doesn't seem to work. :(

Answer 7

we can simplify this radical because
\[z=\frac{8y\pm\sqrt{64y^2-4}}{2}=\frac{8y+2\sqrt{16y^2-1}}{2}\]

Answer 8

the 12 is a mistake

Answer 9

oh my god

Answer 10

i've been on this problem forever!! hahaha thank you so much

Answer 11

we are not done yet though right? you have to take the log

Answer 12

you are solving for x, not 10^x

Answer 13

then it'll be x = log(4y+(16y-1)^(1/2))

right?

Answer 14

so you get
\[x=\log(4y+\sqrt{16y^2-1})\]

Answer 15

damn that 12!

Answer 16

actually that answer doesn't work...

Answer 17

crap why not...

Answer 18

oh wait!

Answer 19

oh wait nothing. hmm. inside the radical is the difference of two squares, not a perfect square

Answer 20

something might be wrong with the program?

Answer 21

let me check with wolfram

Answer 22

ok do you have an option for
\[\pm\]?

Answer 23

maybe that is the problem

Answer 24

i don't, it asks for a comma separate list (if there is more than one answer)

Answer 25

ok then you have to put in both

Answer 26

but the negative in front of the square root should't work right?

Answer 27

oh no that would be ok

Answer 28

4y - (16y^2 - 1) < 0

Answer 29

because
\[16y^2-1<16y^2\] and
\[\sqrt{16y^2}=4y\] so
\[\sqrt{16y^2-1}<4y\] so
\[0<4y-\sqrt{16y^2-1}\]

Answer 30

you really are a legend

Answer 31

in simple english the root is less than 4y, so when you subtract it is still positive

Answer 32

try that and see. if it still doesn't work, i am stuck

Answer 33

confident that the answer is right, but stuck as to how to put it in.
and thanks for the compliment :)

Answer 34

it works. thank so much!

Answer 35

can i ask another question?

Answer 36

sure

Answer 37

(lol sorry)

Answer 38

hope it is an easy one!

Answer 39

farthest i got is...
(where e^(4x) = u)

0= u^2 - yu^2 + y - 1

Answer 40

good god,hold on i will try it

Answer 41

where did you get the "e" from?

Answer 42

it's part of the equation. maybe i don't understand your question

Answer 43

did you see the picture?

Answer 44

(i just substituted for u, to make it easier to type onto here)

Answer 45

the one i see says
\[y=\frac{10^{4x}-10^{4x}}{10^{4x}+10^{4x}}\]

Answer 46

i'm sorry that's a similar question. what i meant to attach was this picture

Answer 47

so far i got to
\[y(10^{4x}+1)=10^{4x}-1\]

Answer 48

do they both work the same way?

Answer 49

(same steps?)

Answer 50

i think

Answer 51

i would get
\[y(e^{4x}-1)=e^{4x}+1\] right?

Answer 52

no that is wrong, should be
\[e^{8x}\] sorry

Answer 53

but how can i solve for x from there?

Answer 54

oh this one is loads easier

Answer 55

ok we have
\[y(e^{8x}-1)=e^{8x}+1\] correct me if i make a mistake

Answer 56

multiply out to get
\[ye^{8x}-y=e^{8x}+1\] then
\[ye^{8x}-e^{8x}=y+1\] factor to get
\[e^{8x}(y-1)=y+1\] divide and get
\[e^{8x}=\frac{y+1}{y-1}\] two more steps and you are done

Answer 57

\[8x=\ln(\frac{y+1}{y-1})\]and so
\[x=\frac{1}{8}\ln(\frac{y+1}{y-1})\] or maybe they want
\[x=\frac{1}{8}(\ln(y+1)-\ln(y-1))\]

Answer 58

check my algebra because it is late, but i think that is right

Answer 59

wow... what was screwing me up was i was trying to make it a quadratic equation. i completely get it.

Answer 60

it's right :)

Answer 61

yeah that is why i said it was easier!

Answer 62

webassign?

Answer 63

ugh yes it is

Answer 64

have fun, and good night

Answer 65

thank you so much! and for staying with me for 20 min

Answer 66

yw

Answer 67

you too, thanks again!