Solve for x in terms of y

Question

anonymous · Answer

$$y=\frac{ 10^{x}+10^{-x} }{ 2 }$$

anonymous · Answer

2y= 10^x+ 1/10^x
2y * 10 ^x= 10^2x + 1
take log both sides
log ( 2y * 10^x) = log (10^2x) ( log a+ log B = Log ab
log (2y)+ log 10^x= 2x  ( log 10 to base 10=1)
log 2y +x= 2x
log 2y= x
2y=e^x
y= e^x/2

richyw · Answer

wait a second. that' cant be right

anonymous · Answer

Um... http://www.wolframalpha.com/input/?i=solve+y%3D%5Cfrac%7B+10%5E%7Bx%7D%2B10%5E%7B-x%7D+%7D%7B+2+%7D+for+x

anonymous · Answer

help please

kira_yamato · Answer

You can substitute z = 10^x, and then do some algebraic manipulation and then complete square

anonymous · Answer

So I got rid of the negative exponent by multiplying by 10^x but I don't know how to proceed after that. I got $$2y(10^{x})=10^{2x}+10$$ but I don't know how to proceed

anonymous · Answer

@Kira_Yamato would that create more work to do at the end b/c you have to reverse the substitution?

kira_yamato · Answer

Maybe... But it seems easier to me

anonymous · Answer

yes I know how to do the substitution way but I was wondering how to solve it this way

kira_yamato · Answer

attachment

anonymous · Answer

THank yOu