# Challenging question on logarithm

• Oct 15th 2009, 06:55 PM
acc100jt
Challenging question on logarithm
Given that $log_{9}x=log_{12}y=log_{16}(x+y)$, find $\frac{x}{y}$

Thanks :)
• Oct 15th 2009, 08:34 PM
PatrickFoster
$z = \frac{y}{x}$

$log_{9}x = log_{12}xz = log_{16}x(1+z)$

$
\frac{lnx}{ln9} = \frac{lnx + lnz}{ln12} = \frac{lnx + ln(1+z)}{ln16}
$

$
lnx + lnx + ln(1+z) = 2(lnx + lnz)
$

$
ln(1+z) = 2 lnz
$

$1+z = z^2$

$z^2 - z - 1 = 0$

$z = \frac{1+\sqrt{5}}{2} = \frac{y}{x}$

$\frac{x}{y} = \frac{2}{1+\sqrt{5}}$

Patrick
• Oct 16th 2009, 05:13 AM
Hello acc100jt
Quote:

Originally Posted by acc100jt
Given that $log_{9}x=log_{12}y=log_{16}(x+y)$, find $\frac{x}{y}$

Thanks :)

In case you don't follow how Patrick went from here:
Quote:

$\frac{lnx}{ln9} = \frac{lnx + lnz}{ln12} = \frac{lnx + ln(1+z)}{ln16}$
to here:
Quote:

$lnx + lnx + ln(1+z) = 2(lnx + lnz)$
... here's an alternative approach, that just uses the fundamental property of a logarithm, namely
$a=\log_bc \iff b^a=c\quad (c>0)$
$\log_{9}x=\log_{12}y=\log_{16}(x+y)=p$, say

$\Rightarrow$
$9^p=x$
$12^p=y\Rightarrow144^p=y^2$
$16^p=(x+y)$
Then, since $9^p \times 16^p=144^p$:
$x(x+y)=y^2$
$\Rightarrow \left(\frac{x}{y}\right)^2+\left(\frac{x}{y}\right )-1=0$

$\Rightarrow \frac{x}{y}=\frac{-1+\sqrt5}{2}\,\Big(=\frac{2}{1+\sqrt5}\Big)$, taking the positive root, since $x>0,y>0$