Question

Help Please.

Answer 1

Express the given quantity as a single logarithm.

\[\frac{ 1 }{ 7 } \ln(x+2)^7+\frac{ 1 }{ 2 }[\ln x-\ln(x^2+3x+2)^2]\]

Answer 2

Little rusty on your log rules? :D

Answer 3

Yea. unfortunately. :/

Is isn't it that
ln+ln = multiplication
ln-ln = Division

Answer 4

yes also,
a*ln = exponent

Answer 5

What does a represent? a constant?

Answer 6

\[\rm \log(a)+\log(b)=\log(ab)\]\[\rm \log(a)-\log(b)=\log\left(\frac{a}{b}\right)\]\[\rm a\cdot \log(b)=\log(b^a)\]

Answer 7

Ya sure

Answer 8

Oh okie

Answer 9

so for the first log we have

(1/7) ln(x + 2)^7 = ln [ (x + 2)^7(1/7 ] = ln (x + 2)

Answer 10

- using the third of the laws

Answer 11

Still having some trouble with this problem from the other day Senor Sid?

Answer 12

Yea, @zepdrix I got it wrong, am still confused... :/

Answer 13

\[\Large\rm \frac17 \ln(x+2)^7\color{red}{+}\frac12\left[\ln x-\ln(x^2+3x+2)^2\right]\]See in red here?
It looks like you accidentally turned it into a multiplication sign somewhere along the way.

Answer 14

Oh there are other problems though :(

Answer 15

Oh but isnt ln+ln=multiplication?

Answer 16

Here is our Log Difference Rule,\[\large\rm \log(a)-\log(b)=\log\left(\frac{a}{b}\right)\]
Notice that the rule is NOT THIS:\[\large\rm \log(a)-\log(b)\ne\frac{\log(a)}{\log(b)}\]Should only be one log after you apply the rule.

Answer 17

Ya maybe we should fix the inside before worrying about that plus.

Answer 18

Oh i see

Answer 19

So the subtraction should change like this,\[\large\rm \ln(x)-\ln(x^2+3x+2)^2\quad=\quad \ln\left[\frac{x}{(x^2+3x+2)^2}\right]\]Ya?

Answer 20

Yea

Answer 21

How bout the 1/2, what can we do with that?\[\large\rm \frac12\ln\left[\frac{x}{(x^2+3x+2)^2}\right]=?\]

Answer 22

Make it into an exponent

Answer 23

So \[\ln[ \frac{ x }{ (x^2+3x+2)^2 }]^{\frac{ 1 }{ 2 }}\]

Answer 24

k great.
And recall that 1/2 power is our square root, ya?
So it might look a little nicer like this,\[\large\rm \ln\sqrt{\frac{x}{(x^2+3x+2)^2}}\]

Answer 25

Yea :D

Answer 26

So far we've simplified our problem down to this,\[\large\rm \frac17\ln(x+2)^7+\ln\sqrt{\frac{x}{(x^2+3x+2)^2}}\]

Answer 27

and the \[\frac{ 1 }{ 7 }\]

Cancels with the exponent 7?

Answer 28

Mmm good.\[\large\rm \ln(x+2)+\ln\sqrt{\frac{x}{(x^2+3x+2)^2}}\]

Answer 29

One more step after that Sid :O
Don't quit there!

Answer 30

i would just expand out the thing first and use the power drop down

ln(x+2) + 1/2 ln(x) - ln[(x+2)(x+1)]

Answer 31

ln(x+2) + (1/2)*ln(x) - ln(x+2) - ln(x+1)

Answer 32

Sorry I was having connection issues.
But what would i do after that?
Multiplication?

Answer 33

\[\ln(x+2) + \frac{ 1 }{2 }[lnx-2\ln(x^2+3x+2)] \]
\[=\ln(x+2) + \frac{ 1 }{ 2 }[lnx-2[\ln(x+1)+\ln(x+2)]]\]
\[=\ln(x+2)+\frac{ 1 }{ 2 }[lnx-2[\ln(x+1)+\ln(x+2)]]\]
\[=\ln(x+2)+\frac{ 1 }{ 2}[lnx-2\ln(x+1)-2\ln(x+2)]\]
\[=\frac{ 1 }{2}lnx-\ln(x-1)\]

Answer 34

Uh I am confused... :/

Answer 35

From where we left off,\[\large\rm \ln(x+2)+\ln\sqrt{\frac{x}{(x^2+3x+2)^2}}\]yes, multiplication,\[\large\rm \ln\left[(x+2)\sqrt{\frac{x}{(x^2+3x+2)^2}}\right]\]But ya, I guess we can simplify the root and square in the denominator,\[\large\rm \ln\left[\frac{(x+2)\sqrt x}{(x^2+3x+2)}\right]\]And from here you can factor as Dan pointed out (maybe we should have done this earlier, but whatever).\[\large\rm \ln\left[\frac{(x+2)\sqrt x}{(x+2)(x+1)}\right]\]Which will simplify a little further, yes?

Answer 36

Oh yea :D

Answer 37

So it would be \[\ln \frac{ \sqrt{x} }{ (x+1) }\]

Answer 38

yes :d

Answer 39

Yes it was the answer. Thanks very much for your help! :)