YAMHT: Yet Another Math Homework Thread

[Bullhonkie]

Text

I've been doing pretty well so far in this class but logarithms always get me.

I can't figure out how to find the x-intercept of #1 algebraically and I'm not quite sure what to do exactly with #3. Feel like I knew how to do the rest but I wouldn't be surprised if it was somehow wrong too.

Any pointers?

 

[youngnhorny]

the answer is 5

 

[Kreon]

looked ok to me, but I didn't really look into it

I too hate logs.
One thing that I've found helps me, at least in the short-term and for checking, is to use graphs to solve/check it

It has backfired on me though

 

[jersiq]

The x intercept occurs when y=0

f(x) is the same as saying y. You should be able to get it from that

 

[Fenixgoon]

you can simplify #3 further.

 

[jamesave]

#3:

(3/2)*6*ln(x)-(3/4)*4*ln(x)

simplify this.

Edit:
this is just a comment.

a*ln(x) = ln ((x)^a)

therefore
ln ((x)^a) = a*ln(x)

 

[Bullhonkie]

Originally posted by: jersiq
The x intercept occurs when y=0

f(x) is the same as saying y. You should be able to get it from that

If you look you can see I tried to go about finding the x intercept in #1 in just that manner, but I kind of lost my way. I know it's supposed to be 3, but I can't quite figure out how to get that.

Originally posted by: Fenixgoon
you can simplify #3 further.

I had it at ln (x^9/x^3) but I wasn't sure if that satisfied the 'single term with ln x' part of the problem. Would it?

 

[2Xtreme21]

Originally posted by: Bullhonkie
Text

I've been doing pretty well so far in this class but logarithms always get me.

I can't figure out how to find the x-intercept of #1 algebraically and I'm not quite sure what to do exactly with #3. Feel like I knew how to do the rest but I wouldn't be surprised if it was somehow wrong too.

Any pointers?

To find the x-intercept, find x when y = 0.

y = (1/2)^(x-3) - 1

1 = (1/2)^(x-3)

log base .5 1 = x-3

(log 1) / (log .5) = x-3

0 = x - 3

x = 3

 

[jamesave]

got it?

 

[Bullhonkie]

Originally posted by: jamesave
#3:

(3/2)*6*ln(x)-(3/4)*4*ln(x)

simplify this.

Doing it this way I get 9 ln(x) - 3 ln(x)
Is that just a case of combining like terms so I end up with 6 ln(x)? Or do I have to apply one of the laws of logarithms?

Originally posted by: 2Xtreme21

log base .5 1 = x-3

(log 1) / (log .5) = x-3

This is the step I don't quite understand. How did we get from the former to the latter?

 

[jersiq]

Originally posted by: Bullhonkie

Originally posted by: jamesave
#3:

(3/2)*6*ln(x)-(3/4)*4*ln(x)

simplify this.

Doing it this way I get 9 ln(x) - 3 ln(x)
Is that just a case of combining like terms so I end up with 6 ln(x)? Or do I have to apply one of the laws of logarithms?

Originally posted by: 2Xtreme21

log base .5 1 = x-3

(log 1) / (log .5) = x-3

This is the step I don't quite understand. How did we get from the former to the latter?

Use the properties of Logarithms
Scroll down past example 1, then in the chart for the properties, look at the last property

Edit: I was replying to your question about the step taken in Question 1

 

[jamesave]

Originally posted by: Bullhonkie

Originally posted by: jamesave
#3:

(3/2)*6*ln(x)-(3/4)*4*ln(x)

simplify this.

Doing it this way I get 9 ln(x) - 3 ln(x)
Is that just a case of combining like terms so I end up with 6 ln(x)? Or do I have to apply one of the laws of logarithms?

yup 6 ln(x).

don't get it too complicated with all the laws of logarithms. Basic math law still applies! for example, 9a-6a = 3a. (where a = ln(x))..

 

[Bullhonkie]

Originally posted by: jersiq

Use the properties of Logarithms
Scroll down past example 1, then in the chart for the properties, look at the last property

Edit: I was replying to your question about the step taken in Question 1

Oh duh I see. I was flipped open to the 'Laws of Logarithms' page in my textbook, but what you showed me the book calls the Change of Base formula which is in a totally separate section so I never even considered that possibility.

Originally posted by: jamesave

Originally posted by: Bullhonkie
Doing it this way I get 9 ln(x) - 3 ln(x)
Is that just a case of combining like terms so I end up with 6 ln(x)? Or do I have to apply one of the laws of logarithms?

yup 6 ln(x).

don't get it too complicated with all the laws of logarithms. Basic math law still applies! for example, 9a-6a = 3a. (where a = ln(x))..

I see. Funny thing is that's actually how I did the problem in my first attempt, but I thought it was wrong since it was so simple.

Thanks for the help everyone.

 

[DrPizza]

It takes a while to appreciate the usefulness of logs (especially natural logs). With time and practice, they become quite easy. It's just a matter of getting used to the rules.