Question

1/16 = 64 ^ 4x - 3

Answer 1

@hartnn lol help me ?

Answer 2

is that 64^(4x-3)
@wendylisette

Answer 3

@gorv yes

Answer 4

is x part of exponent?

Answer 5

@Rahcel_97 yes

Answer 6

Oh, I see.

Log it.

Answer 7

ok then....convert the a16 and 64 in 4 raise the power

Answer 8

ik i have to divide both sides by 64 first but how do i divide 1/16 by 64 ?

Answer 9

better if u multiply

Answer 10

your confusing me

Answer 11

is it necessary to use logarithm here ?

Answer 12

can you hit both sides with the log?

Answer 13

Idk, I would

Answer 14

if u will multiply by 64 both side than 1/16 will become 4 after multiplication

Answer 15

@wendylisette

Answer 16

yeh Ln but i figured that i had to isolate the 4x - 3 by dividing 64 from both sides

Answer 17

no i have to divide 64 @gorv , the inverse remember

Answer 18

I see too.

Answer 19

i neeeed hartnnnn lol

Answer 20

he is here mate

Answer 21

if 16 would have been multiplied by 4x-3, then you would divide.
but here
its 64^(4x-3)

so you better take log on both sides from very first step.

Answer 22

let him explain ....if everyone will explain u will get confuse

Answer 23

ok hold up

Answer 24

and i would have not used logarithm here at all, but since the question belong to log topic, lets use log to solve it

Answer 25

log or ln ?

Answer 26

doesn't matter

Answer 27

i can never tell the difference to when i use which one

Answer 28

when you see a "e" , use natural log, ln

Answer 29

so \[\log_{64} ^ \frac{ 1 }{ 16 } = 4x-3 \] ?

Answer 30

hmmm., yes
continue ?

Answer 31

who me or you ? Lol

Answer 32

just try to simplify left side first

Answer 33

it gives me -.66

Answer 34

lol if i continue you won't learn :P

Answer 35

did you first do
log (1/16) / log 64
?

Answer 36

oh the change of base ?

Answer 37

yes

Answer 38

ok so question , why do i use the formula and when do i know i will use it ?

Answer 39

whats 1/16 ??
can i write it as 16^-1
?

Answer 40

no its a fraction , 1 over 16

Answer 41

and yeh you can

Answer 42

i get the same answer when i dont do the change of base formula though , so does it really matter ?

Answer 43

generally in \( \log_a b \)
if a and b are integers, we think of change of base formula

so,
log 1/16 = log 16^-1 = - log 16
got this ?

Answer 44

-0.66 in fraction form, what is it ?

Answer 45

why cant i leave it as 1/16 ?

Answer 46

and what will you do with log (1/16) then ???
plug it in calculator ? :P

Answer 47

yes

Answer 48

lol

Answer 49

i mean why not ? haha

Answer 50

its a good practice to not use calcy when we can solve a problem without a calcy

Answer 51

i get the same exact answer either way ! lol but thank you i understand the whole log thingy even better

Answer 52

calcy is a good friend !

Answer 53

i agree :)
and just to verify, what final answer you get ?

Answer 54

.583

Answer 55

\(\huge \checkmark \)

Answer 56

yay (:

Answer 57

:D

Answer 58

way of doing it by hand
- log 16 / log 64 = - log 4^2 / log 4^3 = - 2 log 4/ 3 log 4 = - 2/3 = -0.666

Answer 59

no i didnt even read that , it will confuse me like crazy lol

Answer 60

lol haha!