Posted by **Kameesha** on Tuesday, August 14, 2012 at 1:45pm.

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

- College Algebra -
**Steve**, Tuesday, August 14, 2012 at 1:55pm
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^2}2

- College Algebra -
**Jai**, Tuesday, August 14, 2012 at 2:04pm
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~ :)

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