Question

Need step-by step help on this one...not just the answer...need to know how to get there please

log6 3+ log6 8 - log6 4

Answer 1

hello Jenn
we shall work on this

Answer 2

First observe that every log has same base..

Answer 3

i Need steps please...
There are all log6 - are these considered like terms?

Answer 4

yes.. log to the same base are considered as like ones

Answer 5

Good thought, but because the argument of the log is different they are each different numbers.

Answer 6

how do I proceed?

Answer 7

Logk a + logk b = logk ab

Answer 8

They can be combined using the addition property of logs

Answer 9

which is actually multiplication?

Answer 10

If logs are combined by addition then multiply the inner terms
else if subtraction then divide

Answer 11

As helping has explained.\[log_b(a) + log_b(c) = log_b(a\cdot c)\]

Answer 12

do I work with the first 2 tems, and then the last 2 terms? Multiply 3*8, then get log6 24, then divide by the log 6 4?

Answer 13

Yes

Answer 14

You don't divide by the log, you divide the argument

Answer 15

Just divide the inner terms

Answer 16

So then I would get log6 6?

Answer 17

like logk a - logk b = logk a/b

Answer 18

\[log_b(a) + log_b(c) - log_b(d) = log_b(\frac{a\cdot c}{d})\]

Answer 19

That's right. log6(6)

Answer 20

Which is just 1 right?

Answer 21

I can work with formulas, but the instructions that I have, don't offer the ones that the homework questions are asking me to solve! Curious, the writers of these things get paid, why?

Answer 22

You understand this about logs? \[\log_b(b) = 1\]

Answer 23

Jenn,, logk k is 1, log of a number to same base is always 1

Answer 24

if the log base and the number it has next to it are the same, then they equal 1?

Answer 25

Yes absolutely

Answer 26

MUCHAS GRACIAS! That is VERY helpful to know!

Answer 27

Yes, because recall our exponential form:

\[Let\ k = log_b(b) \implies b^k = b \implies k = 1 \implies log_b(b) = 1\]

Answer 28

That again is super helpful. Just wrote it down for future reference!

Answer 29

Also \[log_b(1) =0\]

Answer 30

Because if we do the same thing:
\[Let \ k = log_b(1)\]\[\implies b^k = 1 = b^0\]\[\implies k = 0\]\[\implies log_b(1) = 0\]

Answer 31

you are really a good student to work with,,,

Answer 32

I just posted another one that need your attention please!