Question

Chapter 5:Indices and Logarithms

Answer 1

\[Given~\log _{p}2=x~and~\log _{p}3=y,\]exprees in terms of x and y.

Answer 2

\[\log _{9}\frac{ p }{ 18 }\]

Answer 3

Hmm this is a cool problem :)

Answer 4

\[\Large\rm \log_9\frac{p}{18}=\log_9 p-\log_9 18\]Understand that step?

Answer 5

yup

Answer 6

We can do some cool stuff with the second term, let's leave that alone for now though.
Let's apply change of base to the first term.

Answer 7

The logs rules work "backwards", so you can simplify ("compress"?) log expressions. When they tell you to "simplify" a log expression, this usually means they will have given you lots of log terms, each containing a simple argument, and they want you to combine everything into one log with a complicated argument. "Simplifying" in this context usually means the opposite of "expanding".

Simplify log2(x) + log2(y).
Since these logs have the same base, the addition outside can be turned into multiplication inside:

log2(x) + log2(y) = log2(xy)

The answer is log2(xy)

Answer 8

use this explanation plz

Answer 9

\[\Large\rm \log_9 p=\frac{\log_p p}{\log_p 9}\]

Answer 10

ya that is how u do it

Answer 11

\[1-2y\]

Answer 12

Hmm? 0_o

Answer 13

Applying change of base,
\[\Large\rm \log_9 p-\log_9 18=\frac{\log_p p}{\log_p 9}-\log_9 18\]Then we can rewrite our numerator and denominators as,\[\Large\rm =\frac{1}{\log_p 3\cdot3}-\log_9 18\]And then applying another log rule gives us,\[\Large\rm =\frac{1}{\log_p3+\log_p 3}-\log_9 18\]

Answer 14

Those steps help a lil bit? I don't wanna just go all the way to the find answer, hopefully you can use some of your rules to get there.

Answer 15

\[=\frac{ 1 }{ 2y }-\frac{ \log _{p}18 }{ \log _{p}9 }\]

Answer 16

Ooo nice :o

Answer 17

\[=\frac{ 1 }{ 2y }-\frac{ x+2y }{ 2y }\]

Answer 18

Ooo looks like yer gettin it! :)

Answer 19

\[=\frac{ 1-x-2y }{2y }\]

Answer 20

\[=\frac{ 1 }{ 2y }(1-x-2y)\]

Answer 21

Thnx @zepdrix

Answer 22

most original question of the day