Question

1. Find the domain of the composite function f o g. f(x)= 2/x-3; g(x)= 7/x

2. Solve the following exponential equation. Exact answers only. 2^1-9x = e^2x

Answer 1

f(g(x)) = 2/(g(x)-3) = 2/(7/x)-3) = 2x/(7-3x)

domain would be all reals except x = 7/3, but we have to recall that g(x) is not defined for x=0.

So the domain of fog is all reals except 0, 7/3


2^(1-9x) = e^2x

take log of both sides, recalling that ln(a^b) = b*lna:

(1-9x)ln2 = 2x
ln2 - 9 ln2 * x = 2x
2x + 9 ln2*x = ln2
x = ln2/(2+9 ln2)
or, if you want to be sneaky,

ln2/(ln(e^2) + ln(2^9))
= ln2/ln(512e^2)
= log_512e^22

Answer 2

1.
fog(x) = 2 /[(7/x) - 3]
simplifying,
2x/(7 - 3x)
Domain: all real numbers except 3, 0 and 7/3

2.
2^(1-9x) = e^(2x)
get ln of both sides:
ln 2^(1-9x) = ln e^(2x)
(1-9x)*(ln 2) = 2x
(1-9x)/(2x) = 1/(ln 2)
1/(2x) - 9/2 = 1/(ln 2)
1/(2x) = 1/(ln 2) + 9/2
2x = 1/[1/(ln 2) + 4.5]
x = 1/[2(1/ln 2 + 4.5)]
x = 1/[2/(ln 2) + 9]
x = 1 / [2/(ln 2) + 9(ln 2) / (ln 2)]
x = 1 / [2 + 9(ln 2)] / ln 2
finally,
x = (ln 2)/(2 + (9 ln 2))

hope this helps~ :)