Question

Let 2 = 9^M and 7 = 9^W. What is log subscript 9 of 28? and log subscript 9 of 3.5^1/6 ?

Answer 1

so, \(\large M=\log_9(2)\) and \(\large W=\log_9(7)\)

Answer 2

Could you explain?

Answer 3

What is log∨9 of 28?

Answer 4

you get how i got what M and W equal...?

Answer 5

no not really, could you explain?

Answer 6

it's the definition of logarithm, isn't it...?

if \(5^2=25\) .... then \(\large \log_5(25)=2\)

Answer 7

so did you add the logs to the given information-- how did you get 9 to the power of M as log of 9?

Answer 8

log ⋄ 9∧M = log 2 ?

Answer 9

it is just the definition of logs

if,
\(\Large a^m=n\)

then, \[\Large \log_a (n) =m\]


... but i think we can show it your way too.
start with the original: \(\large 2=9^M\)

then, take log base 9 of both sides:\[\large \log_9(2) = \log_9\left(9^M\right)\]

Answer 10

okay, this is beginning to make more sense--i definitely had trouble with that rule.

Answer 11

\[\large \implies \log_9(2)=M \cdot \log_9(9)\]\[\large \implies \log_9(2)=M \cdot 1\]\[\large \implies \log_9(2)=M \]

Answer 12

okay that makes sense. thanks.

Answer 13

now you have to write the logs you have in terms of logs of 2 and 7 only...

Answer 14

\[\large {\log_9(28)= \log_9(7\cdot4) \\= \log_9(7)+\log_9(4) \\= \log_9(7)+\log_9(2^2)\\= \log_9(7)+2\log_9(2)\\=W+2M }\]

Answer 15

wow that is really clear. great explanation!

Answer 16

you can do the 2nd one yourself ... what do you get

Answer 17

I'm redoing the problem you just showed me for comprehension-- in a few minutes I'll probably be ready for the second equation.

Answer 18

would putting the equation within a radical help?

Answer 19

1/6 log 9 (3.5) ?

Answer 20

yes ... keep going

Answer 21

\[\large {\log_9\left( 3.5^{1/6} \right)\\=\frac 16 \log_9(3.5)\\=??}\]

Answer 22

yes, that's where I am.

Answer 23

3.5 = 7/2 ....

Answer 24

Ha you're right. It was right in front of me.

Answer 25

1/6 log (W/M) ?

Answer 26

1/6 + log W/ log M ?

Answer 27

\[\large {\log_9\left( 3.5^{1/6} \right)\\=\frac 16 \log_9(3.5)\\=\frac 16\left[ \log _9\left( \frac72 \right) \right]}\]

you can only replace with W and M when log 7 and log 2 are separate.

forget about the 1/6 for now...>>> what's the rule for logs of a division???

Answer 28

they subtract.

Answer 29

right ... use that to separate log 7 from log 2
\[\log \left( \frac ab \right) = \log (a) -\log (b)\]

Answer 30

log 9 (7) - log 9 (2) .

Answer 31

\[\large {\frac 16\left[ \log _9\left( \frac72 \right) \right]\\ }\]

\[\Large ={\frac 16\cdot\left[ \log _9(7) -\log _9(2)\right]\\ }\]now you can replace with W and M

Answer 32

1/6 (W-M) . Is that the answer?

Answer 33

\[\Huge \checkmark\]

Answer 34

Thank you so much.

Answer 35

\[\large {W-M \over6}\]

Answer 36

Ah. right.

Answer 37

no problem ... it's just fine as you wrote it too

Answer 38

See you later.