Question

Which of the following is equivalent to log8 (24)?
I put log(8)/log(24), but it was marked wrong. I used the change of base formula to do it. I don't understand this question. Could anyone explain how to do this?

Answer 1

It looks like this\[\log _{8} 24\]
And then my answer was
\[\frac{ \log _{8} }{ \log _{24} }\]

Answer 2

\[FORMULA: \log_{a}b=\frac{ \log_{c} b }{ \log_{c} a } \]
Assuming that the changed base is c=8=>
\[\log_{8} 24=\frac{ \log_{8} 24 }{ \log_{8} 8}\] which is true considering the fact that
\[\log_{8} 8=1\]

Answer 3

So.. it's
\[\frac{ \log 24 }{ \log 8 }\]?
Also, a correction on my above statement. My answer WAS
\[\frac{ \log 8 }{ \log 24 }\]

And there were no logs with the base 8 in any of the answers. \[\log _{8}\]

Did I just flip the formula by accident?

Answer 4

when you write \[\log_{} 24 \] you have a log without base (a nonsens)

Answer 5

Here, this is what the actual question looks like

Answer 6

Wouldn't the log have a base of ten if nothings's written as the base?

Answer 7

no it's not nothing is actually noted lg

Answer 8

but in this circumstances the correct answer is the first one

Answer 9

I can give u the explanation

Answer 10

So basically all of the possible answers are wrong? Am I just learning a really watered down form of logs, maybe? I'm using FLVS, so I wouldn't be surprised

Answer 11

Sure. This base stuff is seriously confusing me.

Answer 12

now pay attention to my explication

Answer 13

you have \[\log_{8} 24\] which can be written with the base change formula as;
\[\log_{8} 24=\frac{ \log_{10}24 }{ \log_{10} 8}\]

Answer 14

they just omitted the 10 from the base using another notation but in principle is right

Answer 15

Okay. Now, how come that automatic ten appears when you do your change of base formula?

Answer 16

\[FORMULA:\log_{a}b=\frac{ \log_{c} b }{ \log_{c} a} \]
in our case we have: a=8(the original base);
b=24(the exponent of the log);
c=10(the changed base);

Answer 17

Why does the base change to ten? I guess part of my confusion come from the fact that I don't quite understand what the base IS. Can you explain that?

Answer 18

You can change in whatever base you like (10,2,5,6,7,8,25,32,.....)
\[\log_{a} b=x=>a^x=b\]

Answer 19

ex:\[\log_{2} 4=2\] because \[2^{2}=4\]

Answer 20

Okay, i think I'm getting this. So, based on what you're saying, then that means that basically, logs are another form of exponential equations, right? But from what I read, logs are supposed to be reflections of exponential equations. Is that wrong?

Answer 21

You're Right

Answer 22

So then, how come \[\log _{a}b=x\] equals \[a^{x}=b\]
Wouldn't they be the inverse of each other?

Answer 23

you mean like\[a^{b}=x\]

Answer 24

The equation you have in the fifth box up.

Answer 25

The idea of logarithms is to reverse the operation of exponentiation, that is raising a number to a power.

Answer 26

How is it reversed? Can you show me an example?

Answer 27

\[Supposing:b^{y}=x=>\log_{b} x=y\]

Answer 28

instead of y you have x in log

Answer 29

Okay, I think I've got it now! It's making sense. Thank you for taking all that time!! I really appreciate it : )

Answer 30

you're welcome