Question

Please help me on a few of these.

1/6 = 64^4x-3

The generation time G for a particular bacteria is the time it takes for the population to double. The bacteria increase in population is shown by the formula G= t/3.3loga^p, where t is the time period of the population increase, a is the number of bacteria at the beginning of the time period, and P is the number of bacteria at the end of the time period. If the generation time for the bacteria is 2 hours, how long will it take 11 of these bacteria to multiply into a colony of 8764 bacteria?

Answer 1

log(9x+5) = 0 solve

Answer 2

log(x+9)-logx=3

Answer 3

log4x+log7 =1

Answer 4

log2x+log12=0

Answer 5

you're using OS for a test? against the code of conduct

Answer 6

its an online assignment that counts as a test grade

Answer 7

than help me?

Answer 8

key word HELP

Answer 9

Which problem do you need help with right now ?

Answer 10

the first one

Answer 11

\(\LARGE\color{blue}{ \bf log_{10}(9x+5) = 0 }\)

when base is not specified it is 10, but in this case it doesn't matter, b/c

\(\LARGE\color{blue}{ \bf log_{~anything~}(1) = 0 }\)

what makes
9x+5=1 ?

Answer 12

@danielle8008 , hello, are you there ?

Answer 13

yes -4/9

Answer 14

Yes, you are correct. Next you need help with number 2 ?

Answer 15

yes please

Answer 16

\(\LARGE\color{blue}{ \bf log(x+9)-log(x)=3 }\)

\(\LARGE\color{blue}{ \bf log(x+9)-log(x)=3~log~10 }\)

\(\LARGE\color{blue}{ \bf log(x+9)-log(x)=log~10^3 }\)

\(\LARGE\color{blue}{ \bf log(x+9)-log(x)=log~1000 }\)

\(\LARGE\color{blue}{ \bf log(x+9)=log~1000 +log~x }\)

\(\LARGE\color{blue}{ \bf log(x+9)=log~1000x }\)

\(\LARGE\color{blue}{ \bf x+9=1000 x }\)

ask anything if you don't get about what I did

Answer 17

hmm ok
the third one now

Answer 18

\(\LARGE\color{red}{rule:~~~~ \bf B~log(A)=log(A^{B})}\)

Answer 19

Ohh, i thought you didn't understand the third equation.... :)

Answer 20

\(\LARGE\color{green}{ \bf log4x+log7 =1 }\)

\(\LARGE\color{green}{ \bf log(4x \times 7 )=1 }\)

\(\LARGE\color{green}{ \bf log(28x )=1 }\)

\(\LARGE\color{green}{ \bf log(28x )=1~log10 }\)

\(\LARGE\color{green}{ \bf log(28x )=log10^1 }\)

\(\LARGE\color{green}{ \bf log(28x )=log10 }\)

\(\LARGE\color{green}{ \bf 28x =10 }\)

Answer 21

on of the answer choices is not 10/28
or 2.8

Answer 22

*one of

Answer 23

are you telling me that 10/28 =2.8 ? Do not confuse 10/28 with 28/10.

Answer 24

one is 280

Answer 25

you got to be kidding me right now.

Answer 26

10/28=5/14 ≈0.36

Answer 27

oh lol

Answer 28

duh sorry

Answer 29

brain fart

Answer 30

so do you still need help with this problem (number 3 ) ??

Answer 31

i understand that one

Answer 32

can you help me with two other ones i did not post?

Answer 33

you actually have 1 left that you posted :)

Answer 34

The number of bacteria present in a culture after t minutes is given as b=100e^kt There are 6639 bacteria present after 5 minutes. Find k. Explain how you solve this problem.

Answer 35

Hold on, you didn't do

\(\LARGE\color{green}{ \bf log2x+log12=0 }\)

Answer 36

oh yeah

Answer 37

\(\LARGE\color{green}{ \bf log2x+log12=0 }\)

\(\LARGE\color{green}{ \bf log2x=-log12 }\)

\(\LARGE\color{green}{ \bf log2x=(-1)~log12 }\)

\(\LARGE\color{green}{ \bf log2x=log(12^{-1}) }\)

\(\LARGE\color{green}{ \bf log2x=log(1/12) }\)

\(\LARGE\color{green}{ \bf 2x=1/12 }\)

Answer 38

thanks

Answer 39

tell me the answer to this problem, before we go on.

Answer 40

.357

Answer 41

What about an exact answer ?

Answer 42

.36 ?

Answer 43

im not sure what you are asking

Answer 44

I am not telling you to round up the decimal, I am asking for the exact x value, in terms of a fraction.

Answer 45

1/12/2

Answer 46

that is not a fraction... or not the type of fraction I want.

Answer 47

.08/2

Answer 48

can we go to the next one now? i chose .36 from the answer options
on the assignment

Answer 49

The number of bacteria present in a culture after t minutes is given as b=100e^kt There are 6639 bacteria present after 5 minutes. Find k. Explain how you solve this problem.

\(\LARGE \color{blue}{ \rm b=100e^{kt} }\)

\(\LARGE \color{blue}{ \rm 6639=100e^{~k\times 5} }\)

\(\LARGE \color{blue}{ \rm 6639=100e^{~5k} }\)

I got to go now. I am tired also. I am suggesting to plug in the approximate 2.7 for e. and solve using log.

Answer 50

thanks