Question

h(x) = 2ln((x-1)/2)+3
Range and domain this function
What is the inverse of the function :)

Answer 1

any takers please

Answer 2

ln((x-1)/2)

Answer 3

if the input is \(\frac{x-1}{2}\) then solve
\[\frac{x-1}{2}>0\] for \(x\)

Answer 4

the range of the log is all real numbers

Answer 5

so x must be greater than 1 and y is an element of all the reals?

Answer 6

yes

Answer 7

you want the inverse solve
\[x=2\ln(\frac{y-1}{2})+3\] for \(y\) in a few steps

Answer 8

only one step is not algebra
subtract 3
divide by 2
not algebra step, write in exponential form
etc

Answer 9

I have got to \(-3/2 = ln(x-1/2)\)

Answer 10

??

Answer 11

i just get so confused with exponentials etc :(

Answer 12

k lets go slow

Answer 13

\[x=2\ln(\frac{y-1}{2})+3\] we solve for \(y\) using almost all algebra

Answer 14

thank you for your patience :)

Answer 15

a) subtract 3 get
\[x-3=2\ln(\frac{y-1}{2})\]

Answer 16

b) divide by 2, get
\[\frac{x-3}{2}=\ln(\frac{y-1}{2})\]

Answer 17

c) the only non algebra step
write in equivalent exponential form , the base of the natural log is \(e\) so in exponential form this is
\[\large e^{\frac{x-3}{2}}=\frac{y-1}{2}\]

Answer 18

d) back to algebra
multiply by 2 get
\[\large 2e^{\frac{x-3}{2}}=y-1\]

Answer 19

finally
e) add 1 get
\[\large 2e^{\frac{x-3}{2}}+1=y\] and that is your inverse

Answer 20

domain x is an element of all the reals and y must be greater than 1?

Answer 21

yes i would say that is right

Answer 22

Can you help me with one more and I will leave you alone LOL

Answer 23

since it is the reverse of the previous one with domain range switched

Answer 24

k

Answer 25

i(x) = 1-3x10^2x+3 Domain and range, plus inverse :)

Answer 26

i can't really read it

Answer 27

looks like
\[1-3x10^{2x}+3\]

Answer 28

but maybe
\[1-3\times 10^{2x+3}\]

Answer 29

\[i(x)=1-3\times10^{2x+3}\]

Answer 30

ok domain as usual is all real numbers
range is \((-\infty,1)\) since \(10^x\) is never negative

Answer 31

so range must be less than 1?

Answer 32

yes

Answer 33

you want the inverse solve
\[x=1-3\times 10^{2y+3}\] for \(y\) again almost all algebra

Answer 34

when you get to
\[-\frac{x-1}{3}=10^{2y+3}\] this time write in equivalent logarithmic form

Answer 35

\[\log_{10} (-\frac{ y-1 }{ 3})=2x+3\]

Answer 36

sorry x and y back to front