Question

log(6)36x
use properties of logarithms to expand the logarithmic expression as much as possible.

Answer 1

another type to try to remember!

Answer 2

Re-write it as a power.

Answer 3

36x^6 ??

Answer 4

wait, scratch that

Answer 5

\[\large \log_b(a) = k \iff b^k = a\]

Answer 6

So here we have:
\[\large \log_6(36x) \iff \]

Answer 7

6^x=36

Answer 8

???

Answer 9

Err sorry. Let's let k = that log.
\[\large \log_6(36x) = k \iff\]

Answer 10

Bah! wait

Answer 11

but where does k jump in from?

Answer 12

I made it up. But it's not going to be helpful

Answer 13

lol

Answer 14

Sorry. I think they want you to see that 36 is a power of the base.

Answer 15

6^2=36 ???

Answer 16

Right

Answer 17

\[\log_b\left(b^2\cdot x\right) = \log_b\left(b^2\right) + log_b(x) \]

Answer 18

Then you can simplify the first log there

Answer 19

\[log_b(b^2) = 2log_b(b) = 2\]

Answer 20

memorizing all these formulas is what gets me... maybe I need a cheat sheet!

Answer 21

it all looks like gibberish to me :(

Answer 22

The cheat sheet is this:
\[log_b(a) = k \iff b^k = a\]

Answer 23

From that rule, (and all the laws of exponents) you can find all the laws of logs.

Answer 24

But it mostly takes practice

Answer 25

my brain is very linguistic vs numeric

Answer 26

so is my answer then 2? it seems too simple

Answer 27

Symbols, not numbers.. my explanations have no numbers ;p

Answer 28

It's not 2. It's 2 + log_6(x)

Answer 29

But hold on, cause you'll need to know this later.

Answer 30

Try this on your linguistic brain..

"The log (base b) of some value a, is the exponent you must use on b in order to get a."

Answer 31

So if I ask you , what's the log (base 6) of 36 ?

Answer 32

one second if you don't mind... got an important call. i'll b back!!

Answer 33

sry... lost my train of thought... business call.

Answer 34

np

Answer 35

okay, so I gravitate back to log(2)

Answer 36

huh?

Answer 37

No, try this I want to know what the log (base 5) of 125 is.

Answer 38

geesh, i dunno. you asked a question up there. this stuff is beyond me it seems!

Answer 39

ok, lemme see

Answer 40

Now remember what I said about what the log is..

"The log (base b) of some value a, is the exponent you must use on b in order to get a."

Answer 41

log(5)125=3

Answer 42

Because?

Answer 43

5 to the power of 3 = 125

Answer 44

Right. 3 is the exponent you must use on 5 (the base) to get 'a' (125)

Answer 45

So now lets go back to our example here..
Lets just imagine that there exists some value (we'll call it k) for which
\[\large\log_6(36x) = k \]

Answer 46

Now, if k is the log of that inside part. that means what?

Answer 47

log(6)36x=2

Answer 48

No, it doesn't. It equals k

Answer 49

ohhhhh, ok

Answer 50

It means that \[\large6^k = 36x\] right?

Answer 51

ok, I see that.

Answer 52

Ok, but we also know that \(36 = 6^2\)

Answer 53

So lets write it that way..
\[\implies 6^k = 6^2x\]

Answer 54

Right?

Answer 55

yes, i see the logic there.

Answer 56

Ok, so now lets pretend we know what k is, but we're still wondering about x

Answer 57

We can 'solve' for x by dividing both sides by \(6^2\)

Answer 58

\[\frac{6^k}{6^2} = x\]

Answer 59

And how do you simplify \(\Large\frac{6^k}{6^2} = x\)

Answer 60

'When you divide powers of the same base you subtract the exponent of the denominator from the exponent of the numerator'

Answer 61

\[\large\implies 6^{k-2} = x\]

And now we take the log base 6 of both sides..
\[\implies k-2 = \log_6(x)\]
Solve for k
\[k = 2+\log_6(x)\]

And since we said originally that \(\large k = \log_6(36x)\) we have:
\[\large \implies \log_6(36x) = 2 + log_6(x)\]

Answer 62

But really this is a very meandery way of simplifying. Once you have more practice with logs (rather than getting answers) you will start to see easier short cuts:

\[\qquad \qquad \log_6(36x)\]\[=\log_6(36) + \log_6(x)\]\[=\log_6(6^2) + log_6(x)\]\[=2 + log_6(x)\]

Answer 63

whoa... it's just overwhelming.

Answer 64

I can't imagine retaking the course again after 16 weeks of brain numbing algebra!

Answer 65

Algebra is good exercise

Answer 66

so, thank you for the help! I am so greatful to all the good helpers here

Answer 67

you're right, exercise for the brain, I felt that way in the beginning but fried with this advanced stuff.

Answer 68

\[\qquad \qquad \log_6(36x)\](The log of a product is the sum of the log of each factor)\[= \log_6(36) + \log_6(x)\]\[=\log_6(6^2) + \log_6(x)\](The log of a power is the product of the exponent and the base of the power)\[=2\log_6(6) + \log_6(x)\](The log(base b) of b is 1)\[=2(1) + \log_6(x) = 2 + \log_6(x)\]

Answer 69

i'm going to print that out :)