1. Logarithmical mess :(

So i need some help with two problems this time and the first one is:

ln910+2ln59ln9

I have already tried to simplify it according to the logarithmical laws but i just can't get it right, maybe i find this so difficult because it's natural logarithms which i am not used to dealing with.

The second one is:
lne3+ln(ee3)+ln1e32

I keep getting the answer 3 but that is apperantly not correct and i honestly can't see why. I figured that since e=e^1/2I would after simplifiyng ee3 get 3/2. I also figured that ln1e32
would equal -3/2. That would mean that the whole thing equals 3...but it's not lol. I could really use some help.

2. Re: Logarithmical mess :(

Let me correct the numbers, they jumped all over the place

Number one is

ln(10/9)+2ln(9/5)-ln9

Number two:

ln(e^3)+ln(sqrt e * e^3)+ln (1/ (e^3/2) )

3. Re: Logarithmical mess :(

What do you mean you are getting a mess? Write out your steps.

4. Re: Logarithmical mess :(

Originally Posted by SlipEternal
What do you mean you are getting a mess? Write out your steps.
On the first one i did this:

ln(10/9)+2ln(9/5)-ln9

ln(10/ (3^2)+ln(9/5)^2-ln(3^2)

After this step im blocked.

Number two i did this:

ln(e^3)+ln(sqrt e * e^3)+ln (1/ (e^3/2) )

3+ ln(e^(1/2) * e^3)+3/2

3+ (3/2) +(3/2) = 3+3=6 (it became 6 this time)

5. Re: Logarithmical mess :(

Originally Posted by Foxtrot
ln(10/ (3^2)+ln(9/5)^2-ln(3^2)

After this step im blocked.
Remember that lnA + lnB = ln (AxB), and lnA - lnB = ln(A/B). So your problem becomes

ln((10/9 x (9/5)^2)/9) = ln(10/(5^2)) = ln(10/25) = ln(2/5).

Originally Posted by Foxtrot
Number two i did this:

ln(e^3)+ln(sqrt e * e^3)+ln (1/ (e^3/2) )

3+ ln(e^(1/2) * e^3)+3/2
That should be 3+ ln(e^(1/2) * e^3)- 3/2. Note the minus sign!

Originally Posted by Foxtrot
3+ (3/2) +(3/2) = 3+3=6 (it became 6 this time)
Should be 3 + 1/2 + 3 - 3/2 = 5.

6. Re: Logarithmical mess :(

Hello, Foxtrot!

$\displaystyle \ln\left(\tfrac{10}{9}\right)+2\ln\left(\tfrac{9}{ 5}\right)- \ln(9)$

We have: .$\displaystyle \ln\left(\tfrac{10}{9}\right) + \ln\left(\tfrac{9}{5}\right)^2 - \ln(9)$

. . . . . . $\displaystyle =\;\ln\left(\tfrac{10}{9}\right) + \ln\left(\tfrac{81}{25}\right) - \ln(9)$

. . . . . . $\displaystyle =\;\ln\left(\frac{\frac{10}{9}\cdot\frac{81}{25}}{ 9}\right) \;=\;\ln\left(\tfrac{2}{5}\right)$

$\displaystyle \ln\left(e^3\right) + \ln\left(\sqrt{e}\cdot e^3\right)+ \ln \left(\frac{1}{e^{\frac{3}{2}}}\right)$

We have: .$\displaystyle \ln(e^3) + \ln\left(e^{\frac{1}{2}}\cdot e^3\right) + \ln\left(\frac{1}{e^{\frac{3}{2}}}\right)$

. . . . . . $\displaystyle =\;\ln(e^3) + \ln\left(e^{\frac{7}{2}}\right) + \ln\left(\frac{1}{e^{\frac{3}{2}}}\right)$

. . . . . . $\displaystyle =\;\ln\left(e^3\cdot e^{\frac{7}{2}}\cdot\frac{1}{e^{\frac{3}{2}}} \right)$

. . . . . . $\displaystyle =\;\ln(e^5) \;=\;5$