# easy log problem, did i get it right?

• Apr 20th 2010, 12:44 AM
murkr
easy log problem, did i get it right?
problem #1 $\displaystyle log_{4} 24 - log_{4} 6$

problem #2 $\displaystyle 10log24 - log4$

im doing these two different ways, i kept getting two different answers, can someone please do these the right way so i can understand the procedure, thank you.

PS: when i type problem 1 in my cal i get the answer 1. but when i do it on paper i get the answer 4, and 1 & 4 are both an answer selection so i dont know which one it is. also, the directions says not to use a calculator though.

the directions say:
use the properties of logarithms to find the exact value of the expression. do not use a calculator.
• Apr 20th 2010, 12:51 AM
How about posting both your attempts..it will be more helpful.. we would be abe to tell you the place you went wrong ..
• Apr 20th 2010, 05:27 PM
murkr
Quote:

How about posting both your attempts..it will be more helpful.. we would be abe to tell you the place you went wrong ..

in the first problem i crossed out the $\displaystyle log_{4}$ because they both had the same base which was 4. then i was left with 26 - 4 but i dont know what to do from there.

and then i couldnt do the second problem at all. sorry my attempts are not that good, i dont know the basics to well
• Apr 20th 2010, 05:34 PM
pickslides
Quote:

Originally Posted by murkr
in the first problem i crossed out the $\displaystyle log_{4}$ because they both had the same base which was 4. then i was left with 26 - 4 but i dont know what to do from there.

and then i couldnt do the second problem at all. sorry my attempts are not that good, i dont know the basics to well

The basics are very important in these problems

$\displaystyle \ln a - \ln b = \ln (a\div b)$

$\displaystyle \log_424- \log_46 = \log_4\frac{24}{6}=\dots$

Also recall

$\displaystyle \log_aa = 1$
• Apr 20th 2010, 05:35 PM
sa-ri-ga-ma
Quote:

Originally Posted by murkr
in the first problem i crossed out the $\displaystyle log_{4}$ because they both had the same base which was 4. then i was left with 26 - 4 but i dont know what to do from there.

and then i couldnt do the second problem at all. sorry my attempts are not that good, i dont know the basics to well

log(a) - log(b) in not equal to (a-b).
Go through the laws of multiplication and division in logarithm.
• Apr 20th 2010, 09:13 PM
murkr
Quote:

Originally Posted by pickslides
The basics are very important in these problems

$\displaystyle \ln a - \ln b = \ln (a\div b)$

$\displaystyle \log_424- \log_46 = \log_4\frac{24}{6}=\dots$

Also recall

$\displaystyle \log_aa = 1$

i can do it on my calulator, i do log(24) / log(4) = 2.29
and log(6) / log(4) = 1.29

i get the answer 1 which is answer A but i dont believe that this is the right answer because in the instructions it says dont use a calculator. can you please do the problem in steps, i find it easier for me to understand it better if i see someone do it. thanks again
• Apr 20th 2010, 09:24 PM
bigwave
use the rule mentioned
Quote:

Originally Posted by murkr
i can do it on my calulator, i do log(24) / log(4) = 2.29
and log(6) / log(4) = 1.29

i get the answer 1 which is answer A but i dont believe that this is the right answer because in the instructions it says dont use a calculator. can you please do the problem in steps, i find it easier for me to understand it better if i see someone do it. thanks again

your calculator asumes a log base of 10 so use the rule mentioned
• Apr 20th 2010, 09:39 PM
murkr
Quote:

Originally Posted by bigwave
your calculator asumes a log base of 10 so use the rule mentioned

im just not getting it im sorry..
when you solve the answer yourself without a cal what do you get?
• Apr 20th 2010, 09:56 PM
bigwave
$\displaystyle \log_424- \log_46 = \log_4\left(\frac{24}{6}\right)=\log_4{4}=1$
• Apr 20th 2010, 10:20 PM
murkr
Quote:

Originally Posted by bigwave
$\displaystyle \log_424- \log_46 = \log_4\left(\frac{24}{6}\right)=\log_4{4}=1$

yeah thats the answer i got from doing it on my cal, for some reason i thought doing it on paper would give me a different answer.

how do i put #3 in a calculator? with #2 i just type in log(24)/log(4) / log(6)/log(4) that equaled 1 for the answer.

but i dont know how to type in #3 can you help me with that? last qustion i promise :o
• Apr 21st 2010, 02:27 AM
Quote:

Originally Posted by murkr
problem #1 $\displaystyle log_{4} 24 - log_{4} 6$

problem #2 $\displaystyle 10log24 - log4$

im doing these two different ways, i kept getting two different answers, can someone please do these the right way so i can understand the procedure, thank you.

PS: when i type problem 1 in my cal i get the answer 1. but when i do it on paper i get the answer 4, and 1 & 4 are both an answer selection so i dont know which one it is. also,the directions says not to use a calculator though.

the directions say:
use the properties of logarithms to find the exact value of the expression. do not use a calculator.

Quote:

Originally Posted by murkr
i can do it on my calulator, i do log(24) / log(4) = 2.29
and log(6) / log(4) = 1.29

i get the answer 1 which is answer A but i dont believe that this is the right answer because in the instructions it says dont use a calculator. can you please do the problem in steps, i find it easier for me to understand it better if i see someone do it. thanks again

Quote:

Originally Posted by murkr
im just not getting it im sorry..
when you solve the answer yourself without a cal what do you get?

Quote:

Originally Posted by murkr
yeah thats the answer i got from doing it on my cal, for some reason i thought doing it on paper would give me a different answer.

how do i put #3 in a calculator
? with #2 i just type in log(24)/log(4) / log(6)/log(4) that equaled 1 for the answer.

but i dont know how to type in #3 can you help me with that? last qustion i promise :o

Your directions say: Don't use a calculator, so first of all