Question

Please wait for image. Help with Q9. Answer is given as 9/2 lg x + lg 27/2. Please show steps!

Answer 1

.5log(x)+log(3/2)
could be wrong..did in hurry

Answer 2

I need to know the steps >< The answer is given as 9/2 lg x + lg 27/2

Answer 3

@stgreen

Answer 4

\[\log_{100}(4x^3)=\frac{1}{2}\lg(4x^3)\]

Answer 5

you can see this by the change of base formula, you get
\[\log_{100}(4x^3)=\frac{\lg(4x^3)}{\lg(100)}=\frac{\lg(4x^3)}{2}\]

Answer 6

you can rewrite this as
\[\log_{100}(4x^3)=\frac{1}{2}\left(\lg(4)+\lg(x^3)\right)=\frac{1}{2}\left(\lg(4)+3\lg(x)\right)\]

Answer 7

is the rest ok?

Answer 8

I'm still not sure how to continue... ><

Answer 9

@satellite73

Answer 10

ok lets start at the top

Answer 11

\[2\lg(3\sqrt{x})-\lg(\frac{4}{3x^2})+\log_{100}(4x^3)\]

Answer 12

we need one base, so we convert \(\log_{100}(4x^3)\) to
\[\frac{1}{2}\lg(4x^3)\]

Answer 13

Does \(\lg\) mean the binary logarithm?

Answer 14

\[2\lg(3\sqrt{x})-\lg(\frac{4}{3x^2})+\frac{1}{2}\lg(4x^3)\]

Answer 15

pretty sure it means log base ten

Answer 16

From googling, lg is an iterated logarithm... I've never used it before.

Answer 17

lol no! http://en.wikipedia.org/wiki/Logarithm#Particular_bases

Answer 18

lg is log base 10, that is what I've been taught :)

Answer 19

Well that's just confusing then.

Answer 20

now some other properties of the log
\[2\lg(3\sqrt{x})-\lg(\frac{4}{3x^2})+\frac{1}{2}\lg(4x^3)\]
\[2\lg(3\sqrt{x})-(\lg(4)-\lg(3x^2))+\frac{1}{2}\lg(4x^3)\]
\[2\lg(3)+\frac{1}{2}\lg(x)-\lg(4)+\lg(3)+2\lg(x)+\frac{1}{2}\lg(2)+\frac{3}{2}\lg(x)\] and more algebra after that
a rather annoying problem

Answer 21

no that is wrong

Answer 22

i made an algebra mistake of course
i forgot to distribute the 2 out front

Answer 23

The book's answer appears to be correct

Answer 24

But it's so tedious to type out :/

Answer 25

yeah i made a tyro mistake and forgot to distribute the 2

Answer 26

it is a bear to type all this out, but careful algebra will do it

Answer 27

@agent0smith Could you write on paper and take a pic and attach then? Really need help ><

Answer 28

Hopefully you can follow it. I didn't show every step, some are shown here, like the change of base formula.

Answer 29

sat has it correct until the last line
\[2\lg(3\sqrt{x})-\lg(\frac{4}{3x^2})+\frac{1}{2}\lg(4x^3)\]
\[2\lg(3\sqrt{x})-(\lg(4)-\lg(3x^2))+\frac{1}{2}\lg(4x^3)\]

\[2\lg(3)+2\cdot\frac{1}{2}\lg(x)-\lg(4)+\lg(3)+2\lg(x)+\frac{1}{2}\lg(2^2)+\frac{3}{2}\lg(x) \]
use \(4= 2^2\)
\[2\lg(3)+\lg(3)+\lg(2)-2\lg(2)+\lg(x)+2\lg(x)+\frac{3}{2}\lg(x)\]
\[3\lg(3)-\lg(2)+\frac{9}{2}\lg(x)\]
\[\lg(3^3/2) + \frac{9}{2}\lg(x)\]