# logarithm of a single quantity

• Aug 2nd 2008, 08:20 PM
Kitty216
logarithm of a single quantity
This problem is on my summer packet for pre cal honors this year.

Write the following expression as a logarithm of a single quantity:
1/2[ln(x+1)+2ln(x-1)-3ln(2x+5)]+3lnx-2

I tried my best to solve it and ended up with this:

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

I'm not sure if the two at the end can sit out like that. It would be great if someone can help =]
• Aug 2nd 2008, 08:25 PM
Jhevon
Quote:

Originally Posted by Kitty216
This problem is on my summer packet for pre cal honors this year.

Write the following expression as a logarithm of a single quantity:
1/2[ln(x+1)+2ln(x-1)-3ln(2x+5)]+3lnx-2

I tried my best to solve it and ended up with this:

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

I'm not sure if the two at the end can sit out like that. It would be great if someone can help =]

almost! you forgot the guy in red
• Aug 2nd 2008, 08:27 PM
Kitty216
o.o oh.. it turned into (x^3) or is that incorrect?

thankies!!
• Aug 2nd 2008, 08:29 PM
Chris L T521
Quote:

Originally Posted by Kitty216
This problem is on my summer packet for pre cal honors this year.

Write the following expression as a logarithm of a single quantity:
1/2[ln(x+1)+2ln(x-1)-3ln(2x+5)]+3lnx-2

I tried my best to solve it and ended up with this:

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

I'm not sure if the two at the end can sit out like that. It would be great if someone can help =]

$\displaystyle \tfrac{1}{2}\left[\ln(x+1)+2\ln(x-1)-3\ln(2x+5)\right]+3\ln(x)-2$

Distribute the $\displaystyle \tfrac{1}{2}$ throughout...

$\displaystyle \tfrac{1}{2}\ln(x+1)+\ln(x-1)-\tfrac{3}{2}\ln(2x+5)+3\ln(x)-2$

Use the power rule of logs...

$\displaystyle \ln(\sqrt{x+1})+\ln(x-1)-\ln((2x+5)^{\tfrac{3}{2}})+\ln(x^3)-2$

Now apply the sum and difference properties of logs:

$\displaystyle \ln(\sqrt{x+1}(x-1)x^3)-\ln(2x+5)^{\frac{3}{2}}-2$

$\displaystyle \implies\ln\left(\frac{\sqrt{x+1}(x-1)x^3}{(2x+5)^{\frac{3}{2}}}\right)-2$

Now how do we bring the 2 into the logarithm?

We know that $\displaystyle \ln(e)=1$, thus $\displaystyle 2\ln(e)=2$.

Therefore, we now have:

$\displaystyle \ln\left(\frac{\sqrt{x+1}(x-1)x^3}{(2x+5)^{\frac{3}{2}}}\right)-2\ln(e)$

$\displaystyle \implies\color{red}\boxed{\ln\left(\frac{\sqrt{x+1 }(x-1)x^3}{e^2(2x+5)^{\frac{3}{2}}}\right)}$

Does this make sense?

I hope this clarifies things!! (Sun)

--Chris

EDIT: Jhevon beat me... (Crying)
• Aug 2nd 2008, 08:40 PM
Kitty216
WOW! I get it! I would never have thought about putting in the ln(e).

Thank you guys!!
• Aug 26th 2008, 07:32 AM
kevin665
logaritem of a single quantity
Hi ,this is Kevin i had completed my Mathematics.and the concepts are very interested.This may be help full for the math students And for the young youth.And they may learn some knowledge from this forum.And i really appreciate for this thought.
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Kevin00

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