Logarithmic equation

**JanClaesen** · Jul3-09, 03:29 PM

log9 (p) = log12 (q) = log16 (p+q)
What is q/p?

9^x + 12^x = 16^x
calculate x
(4/3)^x = q/p

I guess that would work if I'd know how to solve that equation (without using a calculator), unfortunately I don't...

3^(2x)+4^x*3^x = 4^2x

**g_edgar** · Jul3-09, 05:59 PM

Originally Posted by JanClaesen

3^(2x)+4^x*3^x = 4^2x

Divide by 4^x 3^x ... Can you solve a + 1 = 1/a ?

**JanClaesen** · Jul3-09, 08:21 PM

Thanks! :-)

(4/3)^-x + 1 = (4/3)^x

substitute (4/3)^x by y

y^-1 + 1 = y^1

multiply by y

1 + y - y^2 = 0

(1/2) * (1 + sq(5) ) = (4/3)^x = q/p
(1 - sq(5) < 0 , not a possible solution)

**nickto21** · Jul4-09, 05:34 AM

Originally Posted by g_edgar

Divide by 4^x 3^x ... Can you solve a + 1 = 1/a ?

What the heck is a+1=1/a? I can't get an answer. I get to (a squared + a)=1, then I'm stuck.

Steve

**Hootenanny** · Jul4-09, 05:36 AM

Originally Posted by nickto21

What the heck is a+1=1/a? I can't get an answer. I get to (a squared + a)=1, then I'm stuck.

Steve

Try taking a factor of a out of the LHS ...

**g_edgar** · Jul4-09, 08:21 AM

Originally Posted by nickto21

What the heck is a+1=1/a? I can't get an answer. I get to (a squared + a)=1, then I'm stuck.

quadratic equation

~~Дьявол~~ · Jul4-09, 03:53 PM

I do not understand what is the original task. Could you possibly rewrite it?

**JanClaesen** · Jul4-09, 06:23 PM

Sure :)
(p en q > 0) :
log9(p) = log12(q) = log16(p + q) :
What is q/p?

~~Дьявол~~ · Jul5-09, 04:39 AM

If log9*(p)=log12*(q) then 9p=12q and q/p = 9/12

**JanClaesen** · Jul5-09, 05:53 AM

log9(p) = log12(q) = log16(p + q)
is
9^x + 12^x = 16^x
(9, 12 and 16 are the bases eh)

If you solve that equation you get:
q/p = (1/2) * (1 + sq(5) )

I'm wondering, if 9 and 12 were coefficients, would this be the solution?

9 log(p) = 12 log(q)
log(q) = 1 so q = 10
log(p) = 12/9 so p = 10^(12/9)

q/p = 10^(1-12/9) = 10 ^ (-1/3)

**HallsofIvy** · Jul5-09, 06:17 AM

Originally Posted by JanClaesen

log9(p) = log12(q) = log16(p + q)
is
9^x + 12^x = 16^x
(9, 12 and 16 are the bases eh)

If you solve that equation you get:
q/p = (1/2) * (1 + sq(5) )

I'm wondering, if 9 and 12 were coefficients, would this be the solution?

9 log(p) = 12 log(q)
log(q) = 1 so q = 10

What reason would you have to think that log(q)= 1?
If 9 log(p)= 12 log(q) then log(p)/log(q)= 12/9= 4/3. You could then say that log(p)/log(q)= log_q(p)= 4/3 so that p= q^4/3 and p/q = q^4/3. There are many possible values for p/q.

For example, if q= 8, p= 16, then 9log(p)= 9log(16)= 9 log(16)= 9log(2^4)= 9(4)log(2)= 36 log(2)= 12(3)log(2)= 12log(2^3)= 12log(8)= 12log(q). So 9 log(p)= 12 log(q) and p/q= 16/8= 2.

But if q= 27, p= 81, then 9log(p)= 9log(81)= 9log(3^4)= 36log(3)= 12(3)log(3)= 12log(3³= 12 log(27)= 12 log(q). Again 9 log(p)= 12 log(q) but now p/q= 81/27= 3.

q/p = 10^(1-12/9) = 10 ^ (-1/3)

~~Дьявол~~ · Jul5-09, 08:04 AM

If $LaTeX Code: log_9(p)=log_{12}(q)=log_{16}(p+q)$ then

$LaTeX Code: y=(\\frac{4}{3})^x$

$LaTeX Code: y_{1,2}=\\frac{1 \\pm \\sqrt{5}}{2}$

If:

$LaTeX Code: (\\frac{4}{3})^x=\\frac{1 \\pm \\sqrt{5}}{2}$

Then what is x?

Spoiler

$LaTeX Code: x=log_\\frac{4}{3}{\\frac{1 + \\sqrt{5}}{2}}$

You would tell me why $LaTeX Code: y_2=\\frac{1 - \\sqrt{5}}{2}$ is not the solution for $LaTeX Code: (\\frac{4}{3})^x$

Regards.

**JanClaesen** · Jul5-09, 11:33 AM

Because it's a negative number, but I already figured out myself in post #3, thanks anyway ;-)