Question

7+ log_2⁡〖y=3 log_2⁡y 〗

Answer 1

\[7+\log_2(y)=3\log_2(2y)\]?

Answer 2

Simplify 7+ log_2 y =3 log_2 y

Answer 3

ok this is an equation, so probably the question is "solve" not "simplify"

Answer 4

the y is on the same level as the log

Answer 5

it says to simplify on the paper

Answer 6

first subtract
\[log_2(y)\] from both sides to get
\[7=3\log_2(2y)-\log_2(y)\]

Answer 7

no i'm sorry your right it is an equation, i was looking at the wrong question

Answer 8

then use two properties of the log
\[3\log_2(2y)=\log_2((2y)^3)=\log_2(8y^3)\]\]

Answer 9

solve not simplify is right

Answer 10

and also use
\[\log_2(8y^3)-\log_2(y)=\log_2(\frac{8y^3}{y})\]

Answer 11

no we are going to find y

Answer 12

so now we have
\[7=\log_2(8y^2)\]

Answer 13

did you follow so far, because that is the important part

Answer 14

no i'm lost, at the beginning when you subtracted from both sides

Answer 15

oh ok lets go slow

Answer 16

we want to solve for y

Answer 17

so we are going to put everything with a y in it on one side, and the number on the other

Answer 18

that is the reason for subtracting
\[\log_2(y)\] from both sides ok?

Answer 19

alright

Answer 20

so on the left side of the equal sign we have 7

Answer 21

on the right side we have
\[3log_2(2y)-\log_2(y)\]

Answer 22

now our job is to write this as one log. that is write it as
\[\log_2(\text{something})\]

Answer 23

where did the (2y) come from

Answer 24

oh crap i thought you wrote 2y but you did not

Answer 25

sorry. the right side is just
\[3\log_2(y)-\log_2(y)\] yes?

Answer 26

oh ok, yes that's right

Answer 27

whew sorry.

Answer 28

it's fine

Answer 29

now our job is still the same. to write it as a single log. that is write it as
\[\log_2(\text{something})\]

Answer 30

so we have
\[3\log_2(y)=\log_2(y^3)\] is that step clear?

Answer 31

we are using
\[\log(x^n)=n\log(x)\] backwards

Answer 32

clear or no?

Answer 33

?how did you get the (y^3)

Answer 34

ok it is a property of the logs that
\[\log(x^n)=n\log(x)\]

Answer 35

we had
\[3\log_2(y)\]

Answer 36

yes, yes. . .okay i get it, i remember learning that in class

Answer 37

so i bring the 3 up as an exponent

Answer 38

now we are going to use one more property of the log that says
\[log(\frac{a}{b})=\log(a)-\log(b)\]

Answer 39

have you seen that one?

Answer 40

yes i have

Answer 41

ok so the right hand side of the equal sign is now
\[\log_2(y^3)-\log_2(y)\]

Answer 42

which we rewrite as
\[\log_2(\frac{y^3}{y})\]

Answer 43

by that property. now we cancel to get
\[\log_2(y^2)\]

Answer 44

how we doing so far?

Answer 45

because we are almost done!

Answer 46

?what did you cancel

Answer 47

i just canceled in
\[\frac{y^3}{y}\] to get
\[y^2\]

Answer 48

oh okay

Answer 49

thats is i rewrote
\[\log_2(\frac{y^3}{y})=\log_2(y^2)\]

Answer 50

so finally (we are almost done, really) we have
\[7=\log_2(y^2)\]

Answer 51

right

Answer 52

now to find y we rewrite this in "equivalent exponential form" that was the whole point of writing as a single log

Answer 53

\[log_b(y)=x\] same as
\[b^x=y\] so in this case we have
\[\log_2(y^2)=7\] same as
\[y^2=2^7\]

Answer 54

making
\[y=2^{\frac{7}{2}}\] or
\[y=\sqrt{128}\] or even
\[y=8\sqrt{2}\]

Answer 55

btw not to confuse you , but when we had
\[3\log_2(y)-\log_2(y)\] we could have written is as
\[2\log_2(y)=\log_2(y^2)\] we get the same answer

Answer 56

what does "btw" mean

Answer 57

maybe simpler to subtract first and then use one property of the log

Answer 58

by the way

Answer 59

lol, oh okay

Answer 60

what does lol mean?

Answer 61

just kidding lol

Answer 62

in any case we have the answer in many forms, whichever one you like

Answer 63

lol, alright. . .thank you very much!

Answer 64

yw