Maths problems (1 Viewer)

[tam89]

1) Simplify: e^ln(x) - ln(e^x)

is ln(e^x) = x? how?

2) Consider the linear function f(x) = a + bx where "a" is a real number and "b" is a nonzero real number.
a) If the inverse of "f" is "g" such that g(x) = c + dx, find c and d in terms of "a" and "b".
b) Can you find all linear functions which are the same as their own inverse?

Ans:
a) c= -a/b
b) f(x) = a -x OR f(x) = x

Thanks

 

[richz]

1) use log rules drop the x to the front, so xIn(e^1) = x

2) let f(x) = y so y=a+bx
so inverse swap x and y so x=a+by

so y = (x-a)b

so y= x/b -a/b

so c = -a/b and d = 1/b

c) if u swap x for y they produce the same eqns.

 

[tam89]

Hey thanks for your help. Just wondering, how do u know which c and d it belongs to? I mean why c= -a/b and d= 1/b?

Is it because to this equation f(x)= d + cx yea? where inverse f is g

 

[richz]

o u swap y for x right.. and then u make y the subject and then u can work out c and d

 

[tam89]

Thanks mate

Btw... would u also help me with this question below?

Consider the following 'proof" that 1/8 > 1/4.

3 > 2 ---> 3ln(1/2) > 2ln(1/2) ---> ln(1/2)^3 > 1n(1/2)^2 ---> ln(1/8) > ln(1/4) ---> 1/8 > 1/4

Comments?

 

[richz]

log rules ie first one 3In(1/2) = In(1/2)^3 = In(1/8)

 

[tam89]

Oh yea... didn't see that. lol Thanks again

 

[shafqat]

Comment: 1/8 is not greater than 1/4. The proof is wrong because ln (1/2) is negative.

 

[richz]

lol didnt notice that

 

[tam89]

Ah... that's right. lol