Help with these problems would be greatly appreciated:
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
3 answers
Perhaps it is the lateness of the hour but I am stuck on the first one.
1. 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
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
2. log2 x^6 + log2 x^10 = k
6log2x + 10log2x = k
16log2x = k
log2x^16 = k
2^k = x^16
x = 2^(k/16)
6log2x + 10log2x = k
16log2x = k
log2x^16 = k
2^k = x^16
x = 2^(k/16)
0 answers