Question

How do I find the inverse function for this doubling function: y=100(2)^(x/0.32)? Can someone please show me? :'(

Answer 1

This is what I have so far:

Answer 2

y= 100(2)^(x/0.32)
x = 100(2)^(y/0.32)
x/100 = 2^(y/0.32)
logx/100 = (y/0.32)log2
0.32logx/100 = ylog2
(0.32logx/100)/2 = y

HELP! D; The answer is supposed to be y= 0.32log of base 2(x/100) I don't know where I'm going wrong ;'(

@Hero and @TuringTest do you guys know how to do this?

Answer 3

It's not difficult

Answer 4

really? ahh, can you please show me how?

Answer 5

Yeah you made a mistake

Answer 6

Your last step was wrong. I used natural logs rather than log10, but when you get to
0.32[ln(x/100)/ln2] it becomes what you were supposed to get.
You seemed to have just removed the log from the 2.

Answer 7

y= 100(2)^(x/0.32)
x = 100(2)^(y/0.32)
x/100 = 2^(y/0.32)
log(x/100) = (y/0.32)log(2)
log(x/100)/log(2) = y/0.32
(0.32(log(x/100)/log(2))) = y

Answer 8

Parentheses are a little off, but yeah, that's it

Answer 9

wait oops I should have put the log2, anyway the answer is:

Answer 10

so wait why does it say log of base 2... not log of base 10

Answer 11

Should be log[10](2)

Answer 12

I think the answer is equivalent anyways

Answer 13

really? ...? o.O
not: \[y=0.32log _{2}(x/100)\]

Answer 14

I'm pretty sure it is equivalent

Answer 15

ohh.. okay thanks!

Answer 16

They must have used some change of base formula to simplify

Answer 17

http://www.mathwords.com/c/change_of_base_formula.htm

Answer 18

or could it be because:

log[a](b) = log b/log a
and so for this situation, [0.32log(x/100)]/log2 --> 0.32log[2](x/100)?

since log2 could be the log a and log(x/100) could be log b?

Answer 19

they used the change of base formula bro