Question

log7 + log (n - 2) = log 6n
Can someone help? I really REALLY do not understand logarithms

Answer 1

is this exactly like it's written in your book? There aren't any small numbers by the logs??

Answer 2

I don't have a book. I am taking my class online.

Answer 3

oh ok. But there are no small numbers?

Answer 4

But yes. That is it exactly. In fact I copy/pasted.

Answer 5

Ok then we are lucky! Do you know what it means to do 10^n? Doing a log is just doing that backwards

Answer 6

Casey, not really. I'm super duper lost.

Answer 7

ok ok no worries.

Answer 8

Do you know what exponents do?

Answer 9

Yeah. They denote how many times you multiply a number by itself. 5^3 would be 5*5*5

Answer 10

perfect.
So 10^3 is 10*10*10, right?
and 10^n just means we multiply 10 together "n" times, and "n" could be any old number we end up putting in there

Answer 11

Okay.

Answer 12

So now logarithms... Just like the reverse of multiplying is dividing, the reverse of 10 to the power of something (10^n) is the log of that something (log n)

Answer 13

so let's say we want to multiply 10 together a certain number of times to get 10,000... and we want to know how many times we have to multiply 10 together to get to 10,000

Answer 14

can you tell how many times it is?

Answer 15

Okay. 4 times.

Answer 16

exactly.

Answer 17

So we right that like this

Answer 18

\[10^x=10,000\]

Answer 19

and because we know that the log is the reverse of 10^something, we'll do the log of both sides to get x by itself so we can see the answer

Answer 20

\[\log(10^x)=\log(10,000)\]

Answer 21

because the log and the 10^ are reverses, they cancel each other and go away

Answer 22

\[x=\log(10,000)\]

Answer 23

and you can do the log of 10,000 on your calculator and see that it's 4, like you already did

Answer 24

Oh. :o

Answer 25

does that make sense?

Answer 26

Yes. :o So now, how do I add logs together?

Answer 27

ok well... we don't have to! That's the great thing! because 10^something and log cancel out, you can do this for you original question:

Answer 28

Now you lost me.

Answer 29

\[10^{\log(7)}=10^{\log(n-2)}=10^{\log(6n)}\]

Answer 30

just wait, you'll get it.

Answer 31

So they all equal the same thing??

Answer 32

just take 10^ each term... like I did. Then, the 10^ and the log cancel, right? So they go away

Answer 33

and you get \[7+n-2=6n\]... where there are no logs to confuse us and we can just find n with normal old algebra

Answer 34

Oh! :o

Answer 35

make a little bit of sense?

Answer 36

yeah. So n would be 5/7?

Answer 37

make sure you subtract the n from both sides, so the right side should be 5n at that point, then divide by 5

Answer 38

Oh its subtracting. Oops XD so n would equal 1.

Answer 39

perfect. Do you get it? Or do you want to do one more similar problem?

Answer 40

Can we do this one; log(5)16 – log(5)2t = log(5)2??
The numbers in the parenthesis are the little numbers

Answer 41

ah ok I thought we were going to go there. Ok, so no big deal, it's just as easy as the last problem. We just need to know what those numbers mean. Every log has a base. Plain old log is actually \[\log_{10}\]but we use it so often we usually don't write the 10

Answer 42

so remember how \[\log_{10}\]was 10^n backwards? Well, \[\log_{5}\] is just 5^n backwards

Answer 43

Okay. So its always going the be b^n backwards? b for the base

Answer 44

exactly... and we usually call the base b. You're way ahead of me :)

Answer 45

So, knowing that, how could you solve this problem?

Answer 46

5^16 - 5^(2t) = 5^2???

Answer 47

you've got the right idea, but remember that the 5^ and the log(5) cancel. Try one more time?

Answer 48

16-2t=2???

Answer 49

there it is!

Answer 50

So with logs I'm just taking their opposite, getting rid of the base and solving the exponents?

Answer 51

yah, that's it exactly.

Answer 52

Try this one... if you get it I'm sure you've got it under control

Answer 53

\[\log_2(4x)=4\]

Answer 54

hrm.
log(2) is just 2^
so 2^(4x) = 4
but the log and the 2^ cancel
so its just 4x = 4
so x = 1?????
Did I do that right???

Answer 55

@caseyrt

Answer 56

just about. You did the left side exactly right. But remember, what you do to one side you have to do to the other side. so the right side needs to have 2^4, too. What do you get now?

Answer 57

Oh so it should be 4x = 2^4?
then 4x = 16
x=4???

Answer 58

Winner winner! You nailed it!

Answer 59

Okay thank you XDD
You wouldn't mind checking my work on logs in the future would you ??? XDD

Answer 60

no problem. I'm not on here super often, but you can message me on fb if you want. I'm just at /caseyrt

Answer 61

Thank you c:

Answer 62

log(a) + log(b) = log(ab)


log7 + log (n - 2) = log 6n

log(7) + log(n - 2) = log(7(n - 2))

log(7(n - 2)) = log(6n)
log(7n - 14) = log(6n)

7n - 14 = 6n

7n - 6n = 14
(7 - 6)n = 14
n = 14

Answer 63

@SavannahWillett and @caseyrt

any objections?

Answer 64

oh shoot.

Answer 65

Remember to always use the "check method"

Answer 66

wait, what do you mean?

Answer 67

@Hero, what is "check method"?

Answer 68

Check:

n = 14

log(7) + log(14 - 2) = log(6(14))

Answer 69

ok I see what you mean. I thought there might have been a button somewhere I was supposed to use to have someone check me. I feel so bad. I'll message her and fix it. Thank you

Answer 70

That kind of makes sense @Hero