Pls help me : )?

Question

Thanks to everyone who helped me...now i am facing another burden... i have solved 65 out of 70 problems but i am stucked with these problems... thanks to those who can help me.

the inverse of f(x)= 2x^2+4x-1;x greater than or equal to one
the inverse of f(x)=2e^x-3 (how can i solve for x in this fxn)
Solve for x

4^(x+1)+16^(x-3)=2
(log (2x+1) base 3) minus (log x^4 base 3) = 6
(log x-2 base two) minus 2 = (log 3 base 2) minus (log of (log x) base 2)

Answer 1

for f(x)=2e^x-3
to find the inverse you replace y for f(x)
y=2e^x-3
then switch x and y
x=2e^y-3
then solve for y
(x+3)/2=e^y
ln[(x+3)/2]=lne^y lne^y=ylne lne=1 1*y=y
replace y with f^-1(x)=ln[(x+3)/2] f^-1(x)=inverse of f(x)

Answer 2

All these problems are pretty simple, only jugglery oriented based on some concepts, I can only give u some tips because it is you who should do it, and my advise will be, never ask others to do problems in mathematics, because u can be confident only if u do it........................

for the inverse questions, take some y=f(x). , now u have a quadratic in x involing 2x^2 + 4x -1-y = 0. Solve for x, that's the answer.

for the next question, do the same f(x)=y. then we can write
e^x = (y+3)/2. now take log on both sides, that's the answer.

for the questions like solve for x, clearly the nos 4 and 16 give the idea that u have to frame a quadratics equation, so take 4^(x+1) as some t, then we can write 16^(x-3) as 4^2(x+1-4) this t^2/4^2(4), that's it u got a quadratic equation , just solve it.

I have work to do, so try out the other one and think in the same line as before, if u don't get it mail it to address
[e-mail].

Answer 3

4^(x+1)+4(2x-6)=4^1/2
4^(x+1)(2x-6)=1/2
(x+1)(2x-6)=1/2
2(2x^2-4x-6)=1
4x^2-8x-12-1=0
4x^2-8x-13=0
use the quadratic formula to find x

2.log(2x+1)/x^4)=6
(2x+1)/x^4=3^6