Question

can somebody help explain how to evaluate for this logarithm?

10^(log12-log3)

Answer 1

i got up to 10^(log4) and i got stuck. i do not know what to do next?

Answer 2

@fallenangelorchid

Answer 3

>.< I'm better with English, history, and writing sorry.

Answer 4

@phi am i forgetting a property about logarithms? i can't think of anything to do with this

Answer 5

the way you "undo" logs is make them the exponent of their base

assuming log 3 means log base 10 (we is probably true)
then
\[ 10^{\log 3} = 3\]

Answer 6

why is it equal to 3?

Answer 7

you should try to remember this rule
10 * 100 = 1000
or , using exponents
\[ 10^2 \cdot 10^2= 10^3 \]
to remember you add the exponents

Answer 8

yeah, i understand that. makes sense

Answer 9

***why is it 3***

that was an example.
\[ 10^{\log (\text{stuff})} = stuff\]

Answer 10

oh i didn't know that. so it's a rule for when logarithms are found in the exponent of a base?

Answer 11

so with
\[ 10^{\log12-\log3} \]
use the first rule (backwards) to write it as
\[ 10^{\log12} \cdot 10^{-\log3} \]

Answer 12

in other words we are using
\[ 10^{\log12} \cdot 10^{-\log3} = 10^{\log12 + (-\log3)} \]

Answer 13

which is the same rule as
\[10^1 \cdot 10^2= 10^3 \]
(I see I have a typo up above)

but with more complicated things than 1 and 2 for the exponent

Answer 14

ok im following well

Answer 15

now what is
\[ 10^{\log12} \]
?
(see the stuff rule)

Answer 16

12

Answer 17

oh cause you add the exponents if they ahve same base right?

Answer 18

so we have
\[ 10^{\log12-\log3} = 10^{\log12} \cdot 10^{-\log3} \\
= 12 \cdot 10^{-\log3}
\]

* add the exponents if they ahve same base right?**
yes, exactly

Answer 19

one more rule
\[ a^{-b} = \frac{1}{a^b} \]
use that to rewrite
\[ 10^{-\log3} \]

Answer 20

wouldn't it be 10^9 if you add exponents

Answer 21

if it was \(10^{12} \cdot 10^{-3} \), but it isn't

Answer 22

how to add the logs and you got 12?

Answer 23

don't lose the bubble...
think of log(12) as a number (it is a number, and your calculator will give you a decimal value for it)

using the add exponent rule (in reverse) we can say
\[ 10^{\log12-\log3} = 10^{\log12} \cdot 10^{-\log3} \\
= 12 \cdot 10^{-\log3}
\]

now use this rule
a^{-b} = \frac{1}{a^b}
to rewrite the second term

Answer 24

now use this rule\[ a^{-b} = \frac{1}{a^b} \]to rewrite the second term

Answer 25

oh cause 10^(log12) = 12 and 10^(-log3) which is 1/(10^(log3) so 12*(1/3)?

Answer 26

yes

Answer 27

so 4 is the value? !

Answer 28

yes

Answer 29

wow thank you. my teacher never taught me the 'stuff' rule apparently

Answer 30

i would fan you again if i could!! xD

Answer 31

It's not standard usage
but it sticks in my head better.

Answer 32

one more thing. i have a question on logarithms in general. can i ask you in another?

Answer 33

ok

Answer 34

thanks!