Verifying Logarithms Homework: Get Help Here

[Tommzar]

Homework Statement

log(1/2xy^2) = ln x + 3 ln y - ln 2

Homework Equations

The Attempt at a Solution

i have been trying for hours on with next to no results even just an insight on how to go about it would be helpful

 

[Mark44]

Do you know how to combine the terms on the right side into one logarithm expression?

Also, you have used log and ln. Do you intend for these to be different-based logs?

 

[Tommzar]

vaguely yes

and yes well from the criteria sheet that is what it reads

 

[Tommzar]

this is what i got by combining the terms on the right ln(xy^3/2)

 

[Mark44]

vaguely yes

Then you'll need to study the properties of logarithms, and particularly what ln a + ln b equals.

After you do that, we'll help you with the rest.

 

[Tommzar]

i think there is a typo on the original criteria sheet as it reads 3 ln y when combined this = y^3 so I am thinking that it should be 2 ln y

if so i think i have figured it out if not I am still stumped

 

[Mark44]

I don't know anything about your problem sheet, but 3 ln y = ln y³

 

[Tommzar]

so therefore the whole right hand side would = ln(xy^3/2) where as the left = log(1/2xy^2)

so where it says 3 ln y it should read 2 ln y

i think i have got this now thanks for your help mark :)

 

[Mark44]

If you are supposed to verify that the given equation is true for all reasonable values of x and y, the question is not well stated, for two reasons: the 3 ln y vs 2 ln y that you mentioned, and log on one side and ln on the other. These mean different things. If I understand the problem, it should be given with one or the other, but not with both.

 

[Дьявол]

I still do not know what is your equation.
Is it:
$$log(\frac{1}{2xy^2}) = ln x + 3 ln y - ln 2$$

or is it
$$log(\frac{1}{2}*xy^2})=ln x + 3lny - ln2$$

You still need to use the fact that
ln y + ln z = ln (y*z)
and
ln y - ln z = ln(y/z)

 

[Tommzar]

I still do not know what is your equation.
Is it:
$$log(\frac{1}{2xy^2}) = ln x + 3 ln y - ln 2$$

or is it
> $$log(\frac{1}{2}*xy^2})=ln x + 3lny - ln2$$ <

You still need to use the fact that
ln y + ln z = ln (y*z)
and
ln y - ln z = ln(y/z)

The highlighted one is the equation i needed to verify

 

[HallsofIvy]

[itex]ln((1/2)xy^2)= ln(1/2)+ ln(x)+ 2 ln(y)[/tex]. Is that the same as the right side?[/itex]

 

[Tommzar]

right i actually remembered how to do this alls i had to do was expand the left side to show it = the right side