# 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

Staff Emeritus
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.