Question

Hi! I would like to know how to solve: log 3=.5, then log 1/9=? If anyone could help, that be would be splendid! (:

Answer 1

log 3=0.5
change that into exponential form

Answer 2

\[Log_{10}3=0.5\]

Answer 3

I don't know how to do that?

Answer 4

\[Log_{b}a=c\]\[b^c=a\]

Answer 5

Using the formula above
\[10^{0.5}=3\]

Answer 6

I have no clue! :/

Answer 7

Tell me anything you know about this; anything

Answer 8

Honestly I don't know anything, other than that this problem is for a math entrance exam for RU, and I haven't had a math class in well over 2 years...but the problem actually reads log a 3= 0.5, then log a 1/9=?

Answer 9

No problem, we will start from beginning then

Answer 10

Thanks! (:

Answer 11

u should be a fan of me <3 :)

Answer 12

I will be! <3 (;

Answer 13

I will start with my own example

Answer 14

okay.

Answer 15

\[3^x=9\]

Based on your intuition what is x

Answer 16

2?

Answer 17

right, how did you figure that out?

Answer 18

Because 3 to the second power equals 9..

Answer 19

wooo thank u :)

Answer 20

no problem(:

Answer 21

Since it was obivous that 3^2=9 , you were able to figure that out
What if it is 3^x=18

Answer 22

In cases that are not obvious we use log to solve

Answer 23

I don't know..I was thinking it might be 3^3, but that equals 27...so I'm absolutely clueless..

Answer 24

Yes you can't by trial and error

Answer 25

So we use log to solve

Answer 26

okay.

Answer 27

So our example is
3^x=18
we want to know x
To do so we convert to log form

Answer 28

okay.

Answer 29

This is how it converts
\[a^b=c---> Log_a c=b\]

Answer 30

Trying converting our example 3^x=18 using form above

Answer 31

so, do I switch the value placements of b with c then?

Answer 32

3^18=x?

Answer 33

no,
\[Log_3 18=x\]

Answer 34

Okay, I think I got it.

Answer 35

Now, you can do reverse process as well
\[Log_ab=c ---> a^c=b\]

Answer 36

Using form above, try converting
log 3=.5 to exponential form

Answer 37

\[Log_{10} 3=.5\]

Answer 38

where did the 10 come from?

Answer 39

or did you just add that?

Answer 40

When nothing is mentioned, we assumed it is ten.

Answer 41

okay. 10^.5=3?

Answer 42

wait,are you sure you question is right?

Answer 43

you mean my answer?...if that's what you mean, then I have no clue. I'm sorry.

Answer 44

forget the first one, solve the second one

Answer 45

Is what you typed to me before, 3^x=18, converts into Log3 18=x?

Answer 46

That's right

Answer 47

okay...so you want me to try to solve log10 3=.5?

Answer 48

you can't because it is like solving 7=2 does not make sense

Answer 49

So, it's an unsolveable question?

Answer 50

yes, makes no sense

Answer 51

oh, I though I could just convert it the same way i converted it for the first formula, but i guess not/..but why?

Answer 52

because there is no x for us to solve

Answer 53

I see!

Answer 54

second one press Log 1/9 in you calculuator

Answer 55

I'm not sure if I input it in correctly, but I got -.954?

Answer 56

http://www.wolframalpha.com/input/?i=Log%5B1%2F9%5D&t=crmtb01

Answer 57

Oh! And I have a question...so the x would also be considered a?

Answer 58

what on earth?

Answer 59

so it would be -2.2?

Answer 60

no, I mean like in the problem 3^x=18->Log3 18=x, which can be substituted for a?

Answer 61

oops that is not right let me do it correctly

Answer 62

\[\log(\frac{1}{9})=\log(3^{-2})=-2\log(3)=-2\times \frac{1}{2}=-1\]

Answer 63

is it
\[\log_x(3)=\frac{1}{2}\]

Answer 64

that is the solution to
Hi! I would like to know how to solve: log 3=.5, then log 1/9=? If anyone could help, that be would be splendid!

Answer 65

gimmick is to write
\[\frac{1}{9}\] as a power of 3 and then use laws of logarithms. that is all

Answer 66

satellite73: where did the -2 come from? Thanks! (:

Answer 67

don't know what this other business is about, maybe correct but not necessary

Answer 68

he was using properites of log

Answer 69

\[3^{-2}=\frac{1}{3^2}=\frac{1}{9}\]

Answer 70

gotcha!

Answer 71

that is where the -2 came from, recognizing that
\[\frac{1}{9}=3^{-2}\]

Answer 72

so did you get this question from an online thingy?

Answer 73

Yeah...I believe I understand this problem now. Thanks! I appreciate it!! ...um yeah from the RUtgers website for a math placement exam...
:/

Answer 74

typical math teacher question. if
\[\log(2)=.6\] and \[\log(5)=.8\] then
\[\log(20)=?\] and you are supposed to say
\[20=2^2\times 5\] so
\[\log(20)=\log(2^2\times 5)=2\log(2)+\log(5)=2\times .6+.8\] etc

Answer 75

I didn't have quite the best pre-calc teacher in high school, but I guess I could blame it on my own lack of initiative to do well, so I'm re-learning the basics again! :D

Answer 76

or
\[\log(2)=.6\] then
\[\log(\frac{1}{16})=?\] and you are supposed to come up with
\[2^{-4}=\frac{1}{16}\] so
\[\log(\frac{1}{16})=\log(2^{-4})=-4\log(2)=-4\times .6\] andso on

Answer 77

im just trying to figure out what base they meant because log base 10 of 3 does not =5

Answer 78

sorry...I have no clue what that was about...I need to brush up on my logarithms and precalculus mathematical functions and all that... but thanks again for helping me solve the problem! (: The answer is -1

Answer 79

i mean we can solve for the base

Answer 80

it is not a known base. doesn't matter what base is

Answer 81

but what they have isn't true

Answer 82

it bothers me lol
i mean why would they even give that part

Answer 83

it says "IF \[\log(3)=\frac{1}{2}\]

Answer 84

must be some base for which that is true. only wanted to see if you knew to write
\[\frac{1}{9}=3^{-2}\] and use property of log

Answer 85

yes thats untrue
they should said log base x of 3 =1/2
that way it wouldn't be false

Answer 86

i guess they should write
\[\log_b(3)=\frac{1}{2}\] if only to make you happy

Answer 87

yes!
but the part is unnecessary at least it isn't untrue

Answer 88

in fact the base is 9

Answer 89

The problem states: given loga 3=.5, then loga (1/9)=? and the answer is -1

Answer 90

lol

Answer 91

yes that is the question and that is the answer by what i wrote above.

Answer 92

whether we knew the base was 9 or not we still get answer
\[\log_9(3)=\frac{1}{2}\] and
\[\log_9(\frac{1}{9})=-1\]

Answer 93

well that wasn't the question i seen but still you don't need the first part for the second part

Answer 94

Okie dokie! thanks guys! (:

Answer 95

\[\log_9(1)-\log_9(9)=0-1=-1\]
since\[9^0=1and 9^1=9\]

Answer 96

imran, you are supposed to solve for the base in the first equation I think.