i still dunno how to do this qn?

Question

express the following in the form In x= ax+b and find a and b.

(a) x^3 = e^(6x-1)
ans: a=2, b= -1/3

(b) xe^-x)= 2.46
ans: a=1, b=0.9

(c) (xe^x)^2 = 30e^-x
ans:a= -1.5,b=1.7

Answer 1

a) x^3 = e^(6x-1)
lnx^3 = 6x-1
3lnx = 6x-1
lnx = 6x/3 -1/3
a= 6/3 or 2 b= -1/3

b) xe^ -x = 2.46
lne^-x =ln(2.46/x)
-x= ln2.46 -lnx
lnx=x+ln2.46

a= 1 b= ln2.46 or .9

c) (x^2)(e^2x) =30e^-x
(x^2)/30 = (e^-x)/ (e^-2x)
ln(x^2)/30 = lne^(-x-2x)
ln(x^2)/30=lne^-3x
lnx^2-ln30=lne^-3x
2lnx=-3x+ln30
lnx = -(3/2)x +(ln30)/2

a= -3/2 or -1.5 b= (ln30)/2 or 1.7

Answer 2

It might be less confusing if you call ln x = Y
1. Here is #1. Take the natural log of each side to get 3 ln x = (6x-1) . Then 3 Y= 6x-1 and
Y=2x-1/3. a is the coefficient of x and b is the constant in this equation.

2. In this one, taking the log gives you
ln x - x = ln 2.46. Then Y = x + ln 2.46. Again, a is the coefficient of x and b is the constant .