3 ms·

now i have the solution for y in x=10^y here it is:x=10^y subtract 10^y from both the sides x-10^y=0 now subtract x from both the sides -10^y = -x divide

by pencil 16y ago

now i have the solution for y in x=10^y here it is:x=10^y subtract 10^y from both the sides x-10^y=0 now subtract x from both the sides -10^y = -x divide by -1 -10^y/-1 = -x/-1 = 10^y = x take log base 10 both sides log10^y = logx = ylog10=logx y = logx/log10

RiderOfGiraffes 16y ago
Ah, very good. You've gone slightly the long way around, but you've got the right answer. The first part you go from x=10^y to -10^y=-x to 10^y=x. Is it not clear that if x=10^y then 10^y=x? The point is that any time A=B, then B=A. Next, you have ended with y=log(x)/log(10). If the logarithms are base 10, what is log(10)? Can you use that to make your result simpler? Next question: if t=log(u) (where the logarithm is base 10) then what does u equal?
- pencil 16y ago
  i'll be back in 45 min
- pencil 16y ago
  i'am not able to solve t=log(u).not even a single step sorry i'am assuming this would end up in some number called 'e' which i came accross somewhere in the past.please help..i'am not sure what that number means and how to get it
  - RiderOfGiraffes 16y ago
    OK, I really don't know how to help you now using this medium. You claim to have watching the Khan Academy video, but this question is answered in that. The simple rules are these: log(a.b) = log(a) + log(b) hence log(a^4) = log(a.a.a.a) = log(a)+log(a)+log(a)+log(a) = 4.log(a) Therefore if a = b^z then log(a) = log(b^z) so log(a) = z.log(b) This is true for all bases. Then we have log_b(b) = 1 : Log (base b) of b is 1. Hence log_b(b^2) = 2 log_b(b^3) = 3 and so on. In all this the basic rule is this: if x = b^y then log_b(x) = log_b(b^y) = y.log_b(b) = y So if the base is 10, then we have: If a = 10^b then log(a) = log(10^b) = b.log(10) = b.1 = b So now, start with the equation t=log(u), assuming the log is base 10, and use the above rules to change it around. When you do that, see if you can get to saying u equals something. Finally, you need to study sections 1, 2 and 3 of this page: http://en.wikipedia.org/wiki/Logarithm http://en.wikipedia.org/wiki/Logarithm What's on that page will also help you to understand the questions.

by pencil 16y ago

RiderOfGiraffes 16y ago

Ah, very good. You've gone slightly the long way around, but you've got the right answer. The first part you go from x=10^y to -10^y=-x to 10^y=x. Is it not clear that if x=10^y then 10^y=x? The point is that any time A=B, then B=A. Next, you have ended with y=log(x)/log(10). If the logarithms are base 10, what is log(10)? Can you use that to make your result simpler? Next question: if t=log(u) (where the logarithm is base 10) then what does u equal?

pencil 16y ago
i'll be back in 45 min
pencil 16y ago
i'am not able to solve t=log(u).not even a single step sorry i'am assuming this would end up in some number called 'e' which i came accross somewhere in the past.please help..i'am not sure what that number means and how to get it
- RiderOfGiraffes 16y ago
  OK, I really don't know how to help you now using this medium. You claim to have watching the Khan Academy video, but this question is answered in that. The simple rules are these: log(a.b) = log(a) + log(b) hence log(a^4) = log(a.a.a.a) = log(a)+log(a)+log(a)+log(a) = 4.log(a) Therefore if a = b^z then log(a) = log(b^z) so log(a) = z.log(b) This is true for all bases. Then we have log_b(b) = 1 : Log (base b) of b is 1. Hence log_b(b^2) = 2 log_b(b^3) = 3 and so on. In all this the basic rule is this: if x = b^y then log_b(x) = log_b(b^y) = y.log_b(b) = y So if the base is 10, then we have: If a = 10^b then log(a) = log(10^b) = b.log(10) = b.1 = b So now, start with the equation t=log(u), assuming the log is base 10, and use the above rules to change it around. When you do that, see if you can get to saying u equals something. Finally, you need to study sections 1, 2 and 3 of this page: http://en.wikipedia.org/wiki/Logarithm http://en.wikipedia.org/wiki/Logarithm What's on that page will also help you to understand the questions.