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
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
2. log2 x^6 + log2 x^10 = k
6log2x + 10log2x = k
16log2x = k
log2x^16 = k
2^k = x^16
x = 2^(k/16)
Sure! I'd be happy to help you with these problems. Let's go through each one step by step:
1. To find the definite integral of dx/(x(1+ln(x)) from e^6 to 1, we can use integration by parts. Let's break it down:
First, let's simplify the integral:
∫(dx/(x(1+ln(x)))) = ∫(dx/x) - ∫(ln(x)dx/(1+ln(x)))
Now, let's integrate each term separately:
∫(dx/x) is simply ln|x| + C.
For the second term, we can use substitution. Let u = 1 + ln(x), then du = (1/x)dx. Rearranging, dx = xdu. Substituting these values in, we have:
∫(ln(x)dx/(1+ln(x))) = ∫[ln(x)(xdu)/u]
= ∫(ln(u)du)
= u * ln(u) - ∫(u * (1/u) du)
= u * ln(u) - ∫du
= u * ln(u) - u + C
Now let's substitute back u = 1 + ln(x):
= (1 + ln(x)) * ln(1 + ln(x)) - (1 + ln(x)) + C
Now, evaluate the expression at the upper limit (e^6) and subtract the value at the lower limit (1):
∫(dx/(x(1+ln(x)))) from e^6 to 1 = [(1 + ln(e^6)) * ln(1 + ln(e^6)) - (1 + ln(e^6))] - [(1 + ln(1)) * ln(1 + ln(1)) - (1 + ln(1))]
Simplifying further, we have:
= [6 * ln(7) - 7] - ln(2)
Therefore, the definite integral of dx/(x(1+ln(x)) from e^6 to 1 is 6ln(7) - ln(2) - 7.
2. To solve for x in terms of k in the equation log[2,x^6) + log[2,x^10] = k, we can use the property of logarithms that states log(a) + log(b) = log(a * b).
Applying this property, we have:
log[2,x^6) + log[2,x^10] = k
log[2,(x^6 * x^10)] = k
log[2,(x^16)] = k
Now, we can rewrite this equation in exponential form by changing the base to 2:
2^k = x^16
To solve for x, we can take the 16th root of both sides:
(2^k)^(1/16) = x
Simplifying further, we have:
x = 2^(k/16)
Therefore, the solution for x in terms of k is x = 2^(k/16).
3. To solve the equation log base 3(log base 3, x) = -2, we can use the property of logarithms that states log(a,log(b,c)) = log(b,c)/log(b,a).
Applying this property, we have:
log base 3 (log base 3, x) = -2
log base 3, x = 3^(-2)
Now, we can rewrite this equation in exponential form:
3^(log base 3, x) = 3^(-2)
Since the base is the same, we can equate the exponents:
log base 3, x = -2
Now, solve for x by changing the equation into exponential form again:
x = 3^(-2)
Simplifying further, we have:
x = 1/9
Therefore, the solution for x in the equation log base 3 (log base 3, x) = -2 is x = 1/9.
I hope this helps! Let me know if you have any further questions.