Answer:
f^-1 (x) = 1+ln( (x-3)/2)
Step-by-step explanation:
hello :
let f(x) = y so : y = 2e^(x-1) +3
calculate x : e^(x-1) = (y-3)/2
for : y-3 > 0 : x-1 = ln( (y-3)/2) so : x= 1+ln( (y-3)/2)
If(x)= 2e^(x-1) +3 , what is f^-1 (x) = 1+ln( (x-3)/2)
Answer:
I think C
Step-by-step explanation:
Answer:
-4.5 , to - 2.5 , to , - 1.1 , to 0.8, to 9/5
Step-by-step explanation:
Answer:
The only positive integers would be anything less than 3. So 1 or 2 would work
Step-by-step explanation:
If something that is already being raised to a power, gets raised to another power, those powers add up. So if it were
, the exponents would add which would make it
.
In the case of your question, if 'n' was 2 it would be to the power of 6 making it less than to the power of. 1 could also work