help needed with logs: evaluating, solving, etc.

[HayStarr]

OK, so we have started the dreaded logs and ln's. This is so foreign to me.

Does anyone have any good webpage examples to help me out a bit? My professor explains it all, and my brain hurts. The book does not go into enough detail.

Anyway...I only got stuck on a few. Anyone want to take a shot? I have been trying for 4 hours, and I can't think anymore.

First, these are ones I think I did right, but I changed them all after a tutor told me I evaluated them incorrectly.

Evaluate:

1. lne^15

is it: 15lne

2. ln 5th root of 3

is it: 1/5lne

And I have no idea about this one:

3. e^lnx^2 (that is e to the lnx squared)

Solve:

4. 8^2x = 2^2x+2

??? I can't find how to do this one anywhere. I thought maybe, to make the 8 into 2^3, and then 2x * 3=6x, but after I solved it all and got 1/4 for x, I tried it and it did not work.

5. log x = square root of 2
can't find anything on this one either, and it seems so simple. I know I just have my math blinders on, and can't SEE the answer.

6. log(3x+7) + log(x-2) =1

I tried log(3x+7)(x-2)=1 but it didn't work. Maybe I am just tired.

Thank you oh supreme Math Wizards!

Hayley

 

[soroban]

Hello, Hayley!

Your work on #4 is correct ... except for one tiny step.

4) Solve: \(\displaystyle \,8^{2x}\:=\:2^{2x+2}\)

We have: \(\displaystyle \2^3)^{2x} \;= \;2^{2x+2}\)

Then: \(\displaystyle \;2^{6x} \;=\;2^{2x+2}\)

Since the bases are equal, we can equate the exponents:
. . \(\displaystyle 6x\:=\:2x\,+\,2\;\;\Rightarrow\;\;4x\:=\:2\)

Therefore: \(\displaystyle \L\:\fbox{x\:=\:\frac{1}{2}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For #5 and #6, you're expected to know this relationship.

. . \(\displaystyle b^x\:=\:N\;\;\Longleftrightarrow\;\;\log_bN \:=\:x\)

\(\displaystyle 5)\;\log x \:= \:\sqrt{2}\)

We have: \(\displaystyle \:\log_{10}x \:=\:\sqrt{2}\)

. . which becomes: \(\displaystyle \L\:\fbox{x \:=\:10^{\sqrt{2}}}\)

\(\displaystyle 6)\;\log(3x\,+\,7)\,+\,\log(x\,-\,2) \;=\;1\)

Your first step is correct . . .

We have: \(\displaystyle \:\log_{10}\left[(3x,+\,7)(x\,-\,2)\right]\:=\:1\)

. . which becomes: \(\displaystyle \3x\,+\,7)(x\,-\,2) \:=\:10^1\)

Then we have: \(\displaystyle \:3x^2\,+\,x\,-\,14\:=\:10\;\;\Rightarrow\;\;3x^2\,+\,x\,-\,24\:=\:0\)

. . which factors: \(\displaystyle \3x\,-\,8)(x\,+\,3) \:=\:0\)

. . and has roots: \(\displaystyle \:x\:=\:\frac{8}{3},\:\not{-3\;\:}\)


We must discard the -3; it creates logs of negative numbers.

Therefore, the only solution is: \(\displaystyle \L\:\fbox{x\:=\:\frac{8}{3}}\)

 

[Mrspi]

Evaluate:

1. lne^15

is it: 15lne

Yes, but what is ln e? Since e<SUP>1</SUP> = e, ln e = 1.....
15 ln e = 15*1
15

2. ln 5th root of 3

is it: 1/5lne

Was the problem really this? ln 5th root of e ?
If so, then (1/5) ln e would be correct, but again, ln e = 1 and your "final answer" would be 1/5

And I have no idea about this one:

3. e^lnx^2 (that is e to the lnx squared)

Remember that the natural log (ln) of something is the power used on "e" to get that something. "ln x<SUP>2</SUP>" is the exponent you use on "e" to get x<SUP>2</SUP>. So, if you actually raise e to that power, you should get x<SUP>2</SUP>, right?

e<SUP>ln x<SUP>2</SUP></SUP> = x<SUP>2</SUP>

 

[skeeter]

e^ln =

ln(e) =