Homework Help: Log problems

1. Sep 28, 2008

musicfairy

Here's 2 I couldn't get.

8log27

log3x2 = log481/2
Solve for x

For the first one I thought I should solve what's in the exponent first and then solve 8 to that power, but I can't figure out what to do with the exponent.

I'm even more clueless on the second one/ The bases are different, and I can't out what to do with the x2 and 81/2

Can someone please explain this to me? What properties should I use to solve these problems?

I'm supposed to do these without a calculator, and that's what I'm trying to do.

2. Sep 28, 2008

tiny-tim

Hi musicfairy!
Hint: 8 = 23, and (ab)c = … ?
Hint: log48 = … ?

3. Sep 28, 2008

musicfairy

Hi.

So I tried to follow the hints you gave me and came up with these. I might have made a few new properties..

8log27 = 23log27
So 2 and log2 kill each other and kill each other and I end up with 21.

For the second one I got

log481/2 = log4(2/3)

log4(3/2) / log4 1 = 3/2

2log3x = 3/2

lnx/ln3 = 3/4

x/3 = e3/4
x = 3e3/4

This problem is tricky...

4. Sep 28, 2008

tiny-tim

Hi musicfairy!
Goodness … you musicfairies are violent!

"kill each other"?

Hint: 23log27 = (2log27)3 = … ?
Not following any of that

For example, log41 = 0, isn't it?

Answer my original question: log48 = … ?

5. Sep 28, 2008

musicfairy

Well, cancel is a better word. I was only quoting someone from class.

23log27 would equal 73???

log48= 3/2

6. Sep 28, 2008

tiny-tim

good fairy!

Woohoo!

And so log481/2 = … ?

7. Sep 28, 2008

HallsofIvy

No, you are correct that the exponential and logarithm with base 2, because they are inverse functions, "kill each other" but you are handling the exponent incorrectly.
$$8^{log_2(7)}= (2^3)^{log_2(7)}= 2^{3log_2(7)}$$
But that is NOT 3*7. 3 log2(7)= log2(73).
$$2^{3 log_2(7)}= 2^{log_2(7^3)}$$
Now, what is that?

Since 8= 23 and 2= 41/2, it is true that 8= 43/2 but htat means 81/2= 4(3/2)(1/2)= 43/4. And now, log4(81/2= log4(43/4)= 3/4.

??? log4 1= 0. log to any base of 1= 0 because a0= 1 for any positive number a- and you can't divide by 0. It is certainly NOT true that log(a)/log(b)= a/b !

Is this a different problem now? Are you thinking that eln a/ln b= a/b? That is definitely not true! It is far easier to use the basic definition of logarithm: If y= loga(x), then x= ay. If log3(x)= 3/4, then x= what?

8. Sep 28, 2008

musicfairy

log3x2 = 3/4
x2 = 33/4
x = 33/8

Is it right now?

9. Sep 28, 2008

tiny-tim

Your fairy logfather says … yes!

10. Sep 28, 2008

musicfairy

Thanks for all the help and words of wisdom. Now I'll go review log properties.