Question

Write the expression as a single logarithm.

2log3 y-1/4log3 z+5log3 w

Answer 1

3 is always under the log

Answer 2

Is this what it looks like?\[\frac{ 2\log_{3}(y-1) }{ 4\log_{3}z }+5\log_{3}w\]

Answer 3

no no y - 1/4

Answer 4

OK, that's better. lol. Because it wouldn't work otherwise, so I figured something wasn't right....

So its\[2\log_{3}(y-\frac{ 1 }{ 4 })\cdot { 4\log_{3}z }+5\log_{3}w\]right?

Answer 5

there is no 4 after (y- 1/4)

Answer 6

Do you know the properties of logs? Like, you might want to start with:\[\Large a\log_{b}M=\log_{b}M^a\]

Answer 7

Ok, got it.

Answer 8

\[2\log_{3}(y-\frac{ 1 }{ 4 })\cdot { \log_{3}z }+5\log_{3}w\]

Answer 9

yup

Answer 10

i didn't under stand the law

Answer 11

Are you sure the first 2 logs are multiplied in your expression?

Answer 12

yup 100%

Answer 13

\[\Large a\log_{b}M=\log_{b}M^a\] so for example\[\Large 5\log_{2}x=\log_{2}x^2\]

This property lets you bring a coefficient in front of a log, inside the log as an exponent; or in the reverse direction, lets you take an exponent that's inside the log function out in front as a coefficient.

Answer 14

okey how can i but numbers under the log in the calculator

Answer 15

Why would you need a calculator for this problem? You are just supposed to combine everything into one log expression. There is no need for a calculator, because you aren't evaluating anything.

Answer 16

Yah right sorry.

Answer 17

but u said something about first 2 logs are multiplied is that low useful here?

Answer 18

The part I'm having trouble with is the first two logs are a product, and you can't really do anything with that. here is a good summary on the properties of logs:
http://www.purplemath.com/modules/logrules.htm
But there is no property to deal with
\[\log_{b}M\cdot \log_{b}N\]

Answer 19

\[\Large \log_{b}(M\cdot N)=log_{b}M+log_{b}N\]

But that doesn't help here.

Answer 20

aammm so maybe we need some help

Answer 21

@AkashdeepDeb ?

Answer 22

is he good?

Answer 23

\[\Large 2\log_{3}(y-\frac{ 1 }{ 4 })\cdot { \log_{3}z }+5\log_{3}w=\]\[\Large \log_{3}(y-\frac{ 1 }{ 4 })^2\cdot { \log_{3}z }+\log_{3}w^5\]

This is what I mean about that exponent rule. Getting those coefficients inside the log functions. That's a start, because each of the log parts is just the log, not a constant times the log. So you can work with the other properties.

Answer 24

hey u write (y-1/4) it's not like that sorry

Answer 25

2log3y - 1/4

Answer 26

so y is for the first term

Answer 27

Ohhhhhhhh.... so is it\[\Large 2\log_{3}y-\frac{ 1 }{ 4 } { \log_{3}z }+5\log_{3}w\]

Answer 28

yeeeeesss lool

Answer 29

Because that will make a WORLD of difference, lol.

Answer 30

OK THEN... now we can get somewhere. :)

Answer 31

^^

Answer 32

Start with the exponent rule, like I showed you above. I'll get you started:

\[\Large 2\log_{3}y-\frac{ 1 }{ 4 } { \log_{3}z }+5\log_{3}w=\]
\[\Large \log_{3}y^2-.....?\]Can you do the 2nd and 3rd terms?

Answer 33

(actually, I did the 3rd term for you up above, so you only need to do the 2nd really, lol)

Answer 34

log3 y^2 - log3 z^1/4 + log3w^5

Answer 35

yes u did but i understood it (:

Answer 36

Good! Now you'll also need this rule:

\[\Large \log_{b}M-\log_{b}N=\log_{b}\left( \frac{ M }{ N } \right)\]

As well as the one I posted above for sums of logs.

Just follow your order of operations, so you'll move from right to left. E.g., it's as if the original problem read\[\Large (2\log_{3}y-\frac{ 1 }{ 4 } { \log_{3}z })+5\log_{3}w\] so deal with the difference first, THEN the sum

Answer 37

oops *from left to right lol, sorry

Answer 38

okey 1 moment

Answer 39

So it will be like this
log3(y^2/z^1/4)+log3 w^5

Answer 40

Right, so now you've gone from 3 logs down to 2. Now you can take the 2 down to 1, using the rule for the sum of logs = log of a product. :)

Answer 41

log3 y^2/z^1/4 w^5 is that correct ?

Answer 42

Sure is! I'm assuming you mean:

\[\Large \log_{3}\left( { \frac{y^2}{z^{1/4}}\cdot w^5} \right)\]Sometimes it's so hard to tell without the editor, lol

Answer 43

yah i meant that. i have to learn to use it lol

Answer 44

lol it looks so much prettier. :) But if you don't use it, just be generous in your use of parentheses and it makes it easier to decipher. :)

Answer 45

Hhh i will, so is the answer correct

Answer 46

Wow i didn't thought that log things that easy. i learned a lot from u ^_^