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Find log (1 Viewer)

Uncle

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Show how u find:

log4 8 ?
It's like saying:

4 to the power of (what) gives 8?

The (what) is 3/2

But,

The way to solve this because ya calculator probably only has log10 and ln, is to enter on your calculator:

log (8) / log (4)

use either ln or log10 it doesnt matter.

Then you get 3/2

capiche?
 

Herremann

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Show how u find:

log4 8 ?
Uncle's right, I'll just fill in all the gaps if you treat it equating x.

Given log4 8 = x

The inverse function of a logarithm is the exponential function of the same base (in this case the 4). Therefore taking 4 to the power of each side:

4log4 8 = 4x

Therefore: 8 = 4x

Now you can take the logarithm of each side:

log (8) = log (4x)

Using your log rules allows you to take the x out as a factor:

log 8 = x log 4

Therefore:

x = ( log 8 ) / ( log 4 )

x = 3/2

Alternatively, if you are sharp enough to see that 4 cubed gives you 64 and the root of 64 is 8, then you can think your way straight to 43/2, therefore giving you 3/2 directly.

Best Regards
Herremann
 

Woteva636

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its called the change of base rule, i cant find my text book so i cant really explain it, but i thought he others had done it pretty well.
 

Drongoski

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its called the change of base rule, i cant find my text book so i cant really explain it, but i thought he others had done it pretty well.
I'm monitoring; waiting for any other approaches.
 

Trebla

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log4 8 ?
Dunno, why you of all people would ask this...(unless there's something we don't know about? lol)
log48 = log4(4 x 2)
= log44 + log42
= 1 + log440.5
= 1 + 0.5log44
= 1.5
 

Drongoski

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Trebla

There are exercises on logs/indices, even in yr 9 & 10 texts, such as finding: log8 16 and log9 27 and students often find them tricky. I'm curious to find out the various ways people approach such exercises. I'm already seeing quite a number of different ones, including yours and I find that interesting.
 
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addikaye03

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Ok then, get your monkey and his method to evaluate: log736.

Best Regards
Herremann
Monkey: thats easy you noob,

log7 36=x

7^x=36

7^x=7^(ln36/ln7)

therefore x=ln36/ln7 or/ 1.841564442

Me: the monkey's pretty amazing

Monkey: Got anymore wise guy?

Me: Aww now your getting rude
 

Herremann

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...7^x=36

to

7^x=7^(ln36/ln7)...
You've skipped a few steps here that are going to confuse some people.

What I was pointing out was that your method is limited to simple indice-balancing questions. This works but denies students a deeper understanding of what logarithms are all about.

A simple:

7x = 36
log 7x = log 36 [taking logs]
x log 7 = log 36 [Using log laws]
x = log 36 / log 7

is the cleaner and more powerful method rather than muddying the water with equivalence of indices. As I was saying, leave that to the monkeys.

Best Regards
Herremann
 

hermand

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You've skipped a few steps here that are going to confuse some people.

What I was pointing out was that your method is limited to simple indice-balancing questions. This works but denies students a deeper understanding of what logarithms are all about.

A simple:

7x = 36
log 7x = log 36 [taking logs]
x log 7 = log 36 [Using log laws]
x = log 36 / log 7

is the cleaner and more powerful method rather than muddying the water with equivalence of indices. As I was saying, leave that to the monkeys.

Best Regards
Herremann
drongoski said he was looking for different methods wise guy.
 

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