Question

(all are log base 5)
log64-log8/3+log2=log4p
solve for p round to 4 decimal places if ness.

Answer 1

\[\log_5(64)-\log_5({8\over3})+\log_5(2)=\log_5(4p) \]right?

Answer 2

correct

Answer 3

\[=\log_5(64*{3\over8}*2)=\log_5(4p)\to4p=48\to p=12\]

Answer 4

I just used the rules
loga-logb=log(a/b)
and
loga+logb=log(ab)
then raised 5 to each side

Answer 5

okay thanks this really helps.. the mistake i made was instead of multiplying (64*3/8*2) i did this (64-3/8+2) im not really great at math

Answer 6

You just gotta learn the log rules:
\[\log a = \log b=\log(ab)\]\[\log(a^n)=n \log a\]

Answer 7

notice that the base is entirely irrelevant to this question

Answer 8

are the bases only irrelevant to the problem if theyre all the same? and @turintest thats one thing they havent told me in my online class... thanks!

Answer 9

@turingtest. shouldnt it be log a + log b = Log (AB)?

Answer 10

yes you are right.
just making sure you noticed ;)

Answer 11

\[\log_{2}N= 1/4\log_{2}16+ 1/2 \log_{2} 49 \]

Answer 12

thanks!

Answer 13

do i start with putting the exponents back or leave it as is?

Answer 14

Notice we can use \[n \log a=\log(a^n)\]to get closer to having everything under one log sign, so try that first...

Answer 15

yes that is "putting the exponents back" as you call it.

Answer 16

\[\log_{2}16^{1/4}+ \log_{2} 49^{1/2}\]

Answer 17

then i would take the natural log of each?

Answer 18

Whoa, let's just look at the exponential fractions first: \[16^{1/4}\] What does that equal? Hint: Remember that\[ x^{a/b}=\sqrt[b]{x^a}\]

Answer 19

\[\sqrt[4]{16^1} \]=1.091

Answer 20

No you used a calculator improperly or something. Try to apply that rule above to \[49^{1/2}\] first, it's easier.

Answer 21

7?

Answer 22

Exactly, it's a square root in disguise.
That means the other is a fourth root. What is the fourth root of 16?
(it's a whole number)

Answer 23

2

Answer 24

Right, so our expression becomes...

Answer 25

\[\log_{2}N= \log_{2}2+\log_{2}7 \]

Answer 26

perfect, now apply our other rule
\[\log a + \log b = \log (ab)\]

Answer 27

\[\log_{2}N= \log_{2}14 \]

Answer 28

Awesome, almost done!
We just have to realize that, once again, because the bases are the same they don't matter!
When we raise 2 to the power of both sides the logs cancel, so N=?

Answer 29

n=14

Answer 30

You got it, congrads, I've gotta have lunch ;)

Answer 31

lol feel free to eat! thanks so much for helping me and explaining everything. your the most help ive gotten with this class in 2 years (long story) but thanks again! its much appreciated

Answer 32

no prob happy to help!

Answer 33

:) thanks again!