Question

How do I evaluate log927?

Could someone be so kind as to explain how to do this?? I do not have these notes, and I am studying for a test.

Answer 1

@cayssaday do you know the basic formula for logarithm?

Answer 2

Unfortunately not :(

Answer 3

Ok, I will explain :
Suppose you have
\[a=\log_b c\]
It implies
\[b^a=c\]
Suppose you have
\[2=\log_{10} 100\]
it implies
\[10^2=100\]
Do you follow?

Answer 4

Yes!

Answer 5

ok there are some properties of logarithm, you should remember these always
\[\log_a b+\log_a c=\log_a {b\times c}\]
I forgot to mention a is called the base, so if you have
\[\log_2 4=> \text{2 is the base here}\]
\[\log_2 {10} => \text{10 is the base here}\]
do you follow?

Answer 6

Yes I do!

Answer 7

Wait but why is the base of the first one 2 and the second one 10? Shouldn't the second one be 2 as well?

Answer 8

Base can be anything, base should be greater than 0 but not 1
\[base>0 , base\ne 1\]

Answer 9

Oh okay cool! Makes sense!

Answer 10

ok, can you find the value of this?
\[\log _2 8=?\]

Answer 11

I guess I do not quite understand ...

Answer 12

Say the value is x
\[\log_2 8=x\]
it implies
\[8=2^x\]
what will be the power of 2 which is equal to 8?

Answer 13

would it be 4? Oh gosh I am sorry I really am not good at math.

Answer 14

2x 2 x 2=?

Answer 15

6

Answer 16

try again?
\[2\times 2 \times 2=?\]

Answer 17

8 lol

Answer 18

what power of 2 is 8?

Answer 19

ummm 3?

Answer 20

good, so
\[\log_2 8=\text {3}\]

Answer 21

do you follow?

Answer 22

Yes I am sorry for the trouble thank you so much!!!

Answer 23

No problem, let's work on your problem
\[\log _9{27}\]
what's the base here?

Answer 24

9

Answer 25

ok, we will need to find a number which when raised to the power of 9 is 27
Do you have any idea?

Answer 26

Well.. 9 ^1 = 9 and 9 ^2 = 81.... But 9*3=27...

Answer 27

uh, no, 9 is raised to a number...

Answer 28

Here's a question: is 27 a power of 3?

Answer 29

Is it 1.5?

Answer 30

oops @whpalmer4
Cayssaday 9 x 9 x 9=729 not 27
Do you know the square of 27?

Answer 31

yes, it is !

Answer 32

@cayssaday how did you find that it's 1.5?

Answer 33

I was wondering the same thing :-)

Answer 34

the square root of 27 is 5.19 right?

Answer 35

What is the square of 27?
\[27^2=?\]

Answer 36

yes, it is. \[\sqrt{27} = \sqrt{3*9} = 3*\sqrt{3} \approx 3*1.7320508 = 5.196...\]

Answer 37

(9)^x =(9)^(1/2) = 3

Answer 38

Godd Cayssaday, keep going

Answer 39

*good

Answer 40

Idk i just guessed sort of

Answer 41

your approach is correct. According to the log property of addition
\[\log_9(27)=\log_9(3\times 9)=\log_93+\log_99\]
I think you can validate your guess now, can't you?

Answer 42

\(5*5 + 2*5*0.2 + 0.2*0.2 = 25 + 2 +0.04 \approx 27\) if we approximate 5.196^2 by \((5+0.2)^2\)

Well, if you know that \[\sqrt{9} = 3\]and\[3^3=27\]then you can go directly to \[(\sqrt{9})^3 = 27\]or\[((9)^{1/2})^3 = 27\]or\[9^{3/2} = 27\]

Answer 43

@jim_thompson5910 probably has an interesting technique up his sleeve as well..

Answer 44

Here's how you do it using the change of base formula and rule # 3 on this link http://www.purplemath.com/modules/logrules.htm


\[\Large y = \log_{9}(27)\]


\[\Large y = \frac{\log(27)}{\log(9)} \ ... \text{Using Change Of Base Formula}\]


\[\Large y = \frac{\log(3^3)}{\log(3^2)}\]


\[\Large y = \frac{3*\log(3)}{2*\log(3)} \ ... \text{Using Rule 3}\]


\[\Large y = \frac{3*\cancel{\log(3)}}{2*\cancel{\log(3)}}\]


\[\Large y = \frac{3}{2}\]


\[\Large y = 1.5\]



Side Note: the change of base formula is explained here http://www.purplemath.com/modules/logrules5.htm

Answer 45

There's really nothing more than knowing they're multiples of each other. I used basic math and got 1.5 .-.

Answer 46

@cayssaday Do you follow ?

Answer 47

Yes sorry I am taking so long to reply, I am taking notes. I do online school and there isn't really lessons or anything so It's really challenging. You guys are helping SOO much thank you!!!

Answer 48

Evaluate log25125

The answer is 1.5 right?

Answer 49

Yes, because \(\sqrt{25} = 25^{1/2} = 5\) and \(25^1 = 25\), so \(125 = 25*5\) and \(\log_{25} 125 = \log_{25} (25*5) = \log_{25} 25 + \log_{25} 5 = 1+1/2\)
Just like before...

Answer 50

Yes yes awesome it is making sense :D