Question

suppose p and q are positive #'s which log9(p)=log12(q)=log16(p+q)

what is q/p?

Answer 1

\[\log_9(p)=\log_{12}(q)=\log_{16}(p+q)\]?

Answer 2

yes

Answer 3

so maybe we can do something with this:
\[y=\log_9(p) \implies 9^y=p \\ y=\log_{12}(q) \implies 12^{y}=q \\ y=\log_{16}(p+q)=\log_{16}(9^y+12^y) \implies 16^y=9^y+12^y \\....\]
just putting on some first thoughts

Answer 4

should i guess values for y now?

Answer 5

y is hard to guess

Answer 6

Try change of base formula should be easier

Answer 7

i seem to remember one like this \[\log_9(p)=\frac{\log(p)}{2\log(3)}\] for example

Answer 8

Yes, exactly \[\log_9 p = \frac{ \log p }{ \log 9 }\]
\[\log _{12}q = \frac{ \log q }{ \log_{12} }\] and note \[\log_9p = \log_{12} q\] so you have a ratio and you can figure out q/p

Answer 9

\[\huge \log_a x= \frac{ \log_b x }{ \log_b a }\] change of base

Answer 10

i can see that but how does that help me solve for p or q?

Answer 11

oh I think I have something playing with what I had above
\[\frac{p}{q}=\frac{3^y}{4^y} \implies 4^y \frac{p}{q}=3^y \text{ or if squaring both sides } \\ \implies 4^{2y}(\frac{p}{q})^2=3^{2y} \\ \text{ so anyways I want to write } 16^y=9^y+12^y \text{ in terms of } 3^y \text{ and } 4^y \\ \text{ so we have } 4^{2y}=3^{2y}+3^y 4^y\]
write the equation 4^y and p/q using the two equations I just mentioned

Answer 12

you will end up with an equation just in terms of p/q
which you could replace with u
the equation I have is a quadratic

Answer 13

And then once you find p/q
you can just flip it to obtain q/p

Answer 14

so in case you don't see you only have two things to replaced
the 3^(2y) thing and the 3^y

Answer 15

you are writing my equation above in terms of 4^y and p/q
\[3^y=4^y \frac{p}{q} \\ 3^{2y}=4^{2y}(\frac{p}{q})^2\]

Answer 16

Where did you get 3y and 4y?

Answer 17

you mean 3^y and 4^y?

Answer 18

from my equations above:
\[y=\log_9(p) \implies 9^y=p \\ y=\log_{12}(q) \implies 12^{y}=q \\ y=\log_{16}(p+q)=\log_{16}(9^y+12^y) \implies 16^y=9^y+12^y \\....\]
\[\frac{p}{q}=\frac{9^y}{12^y}=\frac{3^{y} 3^{y}}{3^{y} 4^{y}}=\frac{3^y}{4^y}\]

Answer 19

im confused about what to do after we get to 4^2y=3^2y+3^y(4^y)

Answer 20

you make these replacements yet:
\[3^y=4^y \frac{p}{q} \\ 3^{2y}=4^{2y}(\frac{p}{q})^2\]

Answer 21

\[p+q = 16^x = \dfrac{(16*9)^x}{9^x} = \dfrac{12^x12^x}{9^x} = \dfrac{q*q}{p}\]

Answer 22

looks we can't avoid a quadratic...

Answer 23

\[4^{2y}=4^{2y}(\frac{p}{q})^2+4^y \frac{p}{q}4^y \\ 4^{2y}=4^{2y}(\frac{p}{q})^2+4^{2y}\frac{p}{q} \\ 1=(\frac{p}{q})^2+\frac{p}{q}\]

Answer 24

ok i replaced it

Answer 25

@ganeshie8 way is faster getting there

Answer 26

@ganeshie8 can you breifly explain what you did

Answer 27

you put log16(p+q) into exponential form and im not sure what you did after that

Answer 28

he multiply by a pretty one called 9^x/9^x

Answer 29

then rearrange some things to see p and q
where p is 9^x and q is 12^x

Answer 30

@dtan5457
The equation \(\log_b a = \color{red}{x}\) is same as the equaiton \(a=b^{\color{red}{x}}\)

Answer 31

\(a=b^{\color{red}{x}}\)

Notice here, \(\color{red}{x}\) is the exponent.

\(\color{red}{\text{logarithm}} = \color{red}{\text{exponent}}\)

Answer 32

\[\log_{16}(p+q) =x\]
is same as the equation
\[ p+q=16^x\]

Answer 33

yes i get that

Answer 34

\[\log_9(p)=\log_{12}(q)=\log_{16}(p+q)=x\]

From above equation, do we have :
\(p = 9^x\)
\(q=12^x\)
\(p+q=16^x\)
?

Answer 35

yes @myininaya had showed that earlier

Answer 36

Now take a look at my first reply

Answer 37

This one :
\[p+q = 16^x = \dfrac{(16*9)^x}{9^x} = \dfrac{12^x12^x}{9^x} = \dfrac{q*q}{p}\]

Answer 38

i get the last 2 fractions but the first one im not sure what you did

Answer 39

why 16 x 9?

Answer 40

Ahh that is hard to explain. We have introduced a new variable \(x\) in order to eliminate the nasty logs.

We are trying to eliminate \(x\) too and get an equation in terms of \(p,q\) only.

Answer 41

In pursuit of that, we're trying to express 16 in terms of 9 and 12

Answer 42

Notice \(16=4*4 = \dfrac{4*4*3*3}{3*3} = \dfrac{12*12}{9}\)

Answer 43

theres no way i would of thought of that but yeah now i see what you did

Answer 44

now that we know q^2/p is equal to 16^x what do we do

Answer 45

That is not just the way to do it. There are multiple ways to do this. You just use whatever the idea that occurs to you

Answer 46

\[p+q = 16^x = \dfrac{(16*9)^x}{9^x} = \dfrac{12^x12^x}{9^x} = \dfrac{q*q}{p} \]

So we have
\[p+q = \dfrac{q^2}{p}\]

Answer 47

logs and x are gone

Answer 48

divide \(q\) through out and let \(\frac{q}{p}=t\)

Answer 49

(p/q)+1=q/p?

Answer 50

or t+1=q/p

Answer 51

(p/q)+1 = q/p

lettinh q/p = t, we get

1/t + 1 = t

Answer 52

which is same as

1 + t = t^2

you can solve it

Answer 53

(1+sqrt5)/2?

Answer 54

the minus is rejected yes?