Question

Please help! Problem dealing with logarithms. Click attached

Answer 1

You will need to use a few log rules for this.
1. logx-logy=log(x/y)
2. zlogx=log(x^z)

Answer 2

yeah for this one only these two rules are needed.
First try to make all the expressions in the form of log(something)-log(something else)....
So get rid of the factors 3, 1/4, 1/2 using the 2. rule

Answer 3

If thats done you can use the first rule.
Good luck!
And remember x^(1/2) is equivalent to the square root of x

Answer 4

Thanks!

Answer 5

:/ Hmm... I think it's either the second or third but having trouble coming up with the correct final answer

Answer 6

Let look at it term by term
3logx=?

Answer 7

1

Answer 8

1? we dont know what x is, so the solution has to have x in it

Answer 9

x=1?

Answer 10

by rule2 3logx=log(x^3)

Answer 11

we dont know what x is, just imagine any number, but you cannot place!!!! any number there instead of x

Answer 12

that is called unknown or variable

Answer 13

so second part 1/4log y=?

Answer 14

(y is also unknown)

Answer 15

Yes. But why am I getting y=1? http://www.wolframalpha.com/input/?i=1%2F4logy%3D

Answer 16

that is its root, different thing
It means that when y=1 1/4log y=0
That is called the root of a function, where its value is 0.

Answer 17

O.k. Now I get it

Answer 18

wolfram is a good tool but you still need to think

Answer 19

thats why we have a magical brain! :-)

Answer 20

:)

Answer 21

here your only goal is to simplify the log expressions into one log expression

Answer 22

you are not solving anything, it will be still the same thing just written differently. Some would say in a neater form. (I disagree but that doesnt matter now I guess)

Answer 23

so 1/4log y=?

Answer 24

logy/4?

Answer 25

that is wrong but better than 1 :)
rule 2 was:
zlogx=log(x^z)
In our scenario z is given, it is 1/4 and x should be replaced by y

Answer 26

log(y^1/4) ?

Answer 27

Yes, you are getting there!

Answer 28

1/2logz=?

Answer 29

1/2log1/4=-/301?

Answer 30

?

Answer 31

1/2log1/4 where did this expression come from? Its like a rabbit from a hat

Answer 32

1/2logz=? where z is unknown, so z should not be replaced by a number!

Answer 33

I can see that these logs are killing you :D, I dont like them as well....

Answer 34

AHHHHHH! Yes they are lol Do you mind just going through solving the problem step by step? I'll follow along and can use it as an example

Answer 35

sure
1/2logz=log(z^1/2)

Answer 36

so we have
3logx=log(x^3)
1/4logy=log(y^1/4)
1/2logz=log(z^1/2)

Answer 37

now first two parts
logx^3 -logy^1/4 (1. rule)=
=log(x^3 /y^1/4)

Answer 38

log(x^3 /y^1/4) -logz^1/2= (1. rule again)
=log){(x^3 /y^1/4) /z^1/2} simplify the middle part now
x^3 /y^1/4) /z^1/2= (fraction over fraction is = to 1st fraction multiplied by the reciprocal of the 2nd fraction)
=(x^3 /y^1/4) *z^1/2=(x^3*z^1/2)/y^1/4

Answer 39

so the solution is
log(x^3*z^1/2)/y^1/4

Answer 40

A

Answer 41

Thank you so very much Andras :) I really appreciate your help and patience haha

Answer 42

Your welcome, hope it cleared up a bit