# Thread: natural logarithim of a single quantitiy

1. ## natural logarithim of a single quantitiy

stuck part way through on simplifying this problem....

2/3[5ln(x+1) - 2ln(x-1) + 4ln(3x-1)] - 2ln(x+5) + 1

2. Originally Posted by RosieLaird
stuck part way through on simplifying this problem....

2/3[5ln(x+1) - 2ln(x-1) + 4ln(3x-1)] - 2ln(x+5) + 1
1. Parentheses do not match

2. You know that $A \ln(B)=\ln(B^A)$ that $\ln(U)+\ln(V)=\ln(UV)$ and $\ln(U)-\ln(V)=\ln(U/V)$, so tell us what problem you are having applying these laws to this simplification.

CB

3. Also keep in mind that $ln|e| = 1$.

4. Hello, RosieLaird!

This cannot be simplified to a single log.

$\frac{2}{3}\bigg[\sin(x+1) - 2\ln(x-1) + 4\ln(3x-1)\bigg] - 2\ln(x+5) + 1$

$\frac{2}{3}\sin(x+1) - \frac{4}{3}\ln(x-1) + \frac{8}{3}\ln(3x-1) - 2\ln(x+3) + 1$

. . $=\;\frac{8}{3}\ln(3x-1) - \frac{4}{3}\ln(x-1) - 2\ln(x+3) + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\frac{2}{3}\bigg[4\ln(3x-1) - 2\ln(x-1) - 3\lnm(x+3)\bigg] + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\frac{2}{3}\bigg[\ln(3x-1)^4 - \ln(x-1)^2 - \ln(x+3)^3\bigg] + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\frac{2}{3}\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right] + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right]^{\frac{2}{3}} + \frac{2}{3}\sin(x+1) + 1$

5. Originally Posted by Soroban
Hello, RosieLaird!

This cannot be simplified to a single log.

$\frac{2}{3}\sin(x+1) - \frac{4}{3}\ln(x-1) + \frac{8}{3}\ln(3x-1) - 2\ln(x+3) + 1$

. . $=\;\frac{8}{3}\ln(3x-1) - \frac{4}{3}\ln(x-1) - 2\ln(x+3) + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\frac{2}{3}\bigg[4\ln(3x-1) - 2\ln(x-1) - 3\lnm(x+3)\bigg] + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\frac{2}{3}\bigg[\ln(3x-1)^4 - \ln(x-1)^2 - \ln(x+3)^3\bigg] + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\frac{2}{3}\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right] + \frac{2}{3}\sin(x+1) + 1$

. . $=\;\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right]^{\frac{2}{3}} + \frac{2}{3}\sin(x+1) + 1$

Just as I misread the parentheses you have misread the first term in the square brackets; it id $5\ln(x+1)$ not $\sin(x+1)$

CB

6. Ha!

That's a new blunder for me . . .

. . Reading 5ln(x+1) to be sin(x+1)

Sheesh, that's embarrassing . . . Hope no one saw it.
.

7. lol I see there this :

$\displaystyle \frac {2}{3[5\ln(x+1)-2\ln(x-1)+4\ln(3x-1)]} -2\ln(x+5)+1$

heheheh confusing

P.S. at first glance that 5ln is sin hehehehe