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September 16, 2014

September 16, 2014

Posted by **Angelina** on Friday, January 30, 2009 at 2:37am.

1.Find the definite integral of dx/(x(1+ln(x))from e^6 to 1.

2. Solve for x in terms of k for log[2,x^6)+log[2,x^10=k. (its log base 2)

3.Solve log base 3(log base 3, x)=-2

- Calculus - beyond me -
**Damon**, Friday, January 30, 2009 at 3:13amPerhaps it is the lateness of the hour but I am stuck on the first one.

- Math -
**drwls**, Friday, January 30, 2009 at 4:13am1. For the indefinite integral, try the substitution ln x = u

e^u = x

e^u du = dx

indef. integral of dx/(x(1+ln(x))

= integral of e^u du/[e^u*(1 + u)

= integral of du/(1+u)

= ln (1 + u) = ln (1 + ln x)

When x = e^6, the indef. integral is ln (1 + 6) = ln 7 = 1.9459

When x = 1, the indef. integral is ln 1 = 0

Def. integral = -ln 7

3. Rewrite as

3^-2 = log3,x = 1/9

3^(1/9) = x

x = 1.129831

- Math -
**Reiny**, Friday, January 30, 2009 at 10:12am2. log

_{2}x^6 + log_{2}x^10 = k

6log_{2}x + 10log_{2}x = k

16log_{2}x = k

log_{2}x^16 = k

2^k = x^16

x = 2^(k/16)

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