solution AS1

# solution AS1 - 1 Solution 9 x = 2 e x 2 x ln2 = x 2 ln 2 x...

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Page 1 Math151 Assignment 1 Solutions - Section 1.5

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Math 151 - Assignment 1 Instructor’s Questions - Solutions 1. Simplify the following expressions completely: (a) e ln(1 / 2) Solution: Since f ( x ) = e x is the inverse function of g ( x ) = ln x we have that 1 2 = f ± g ± 1 2 ¶¶ = e ln(1 / 2) . (b) 2ln( e x ) + 3 ln y e Solution: From ln( e x ) = x and ln( y e ) = e · ln y we get 2ln( e x ) + 3 ln y e = 2 x + 3 e · ln y where x R and y R + . 2. Solve for x using logarithms: (a) 2 e 3 x = 4 e 5 x Solution: 2 e 3 x = 4 e 5 x ln ( 2 e 3 x ) = ln ( 4 e 5 x ) ln2 + ln3 x = ln4 + ln e 5 x ln2 + 3 x = ln4 + 5 x ln2 - ln4 = 5 x - 3 x ln 2 4 = 2 x ln 1 2 = 2 x - ln2 = 2 x x = - ln2 2 (b) 9 x = 2 e x 2
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Unformatted text preview: 1 Solution: 9 x = 2 e x 2 x ln2 = x 2 + ln 2 x 2-x ln9 + ln 2 = 0 x 1 , 2 = ln9 + p (ln9) 2-4ln 2 2 We note that (ln9) 2-4ln 2 &amp;gt; 0, so the given equation has two real solutions. 3. Find the inverse function if (a) f ( t ) = (1 . 04) t Solution: f-1 ( t ) = log 1 . 04 t (b) g ( x ) = 1 + ln x Solution: We write y = 1 + ln x . To nd the inverse function we need to solve x = 1 + ln y,y &amp;gt; for y . Thus x = 1 + ln y,y &amp;gt; ln y = x-1 ,y &amp;gt; y = e x-1 . Thus g-1 ( x ) = e x-1 . 2...
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solution AS1 - 1 Solution 9 x = 2 e x 2 x ln2 = x 2 ln 2 x...

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