Please shed some light on the following......

If x and y are both positive and log_9 x = log_12 y = log_16 (x + y), find the value of x/y.

Note: log_9 x should read as log base 9 x

Thanks.

Please shed some light on the following......

If x and y are both positive and log_9 x = log_12 y = log_16 (x + y), find the value of x/y.

Note: log_9 x should read as log base 9 x

Thanks.

Put:

$u=\log_9(x)=\log_{12}(y)=\log_{16}(x+y)$

Then:

$9^u=x,\ 12^u=y,\ 16^u=x+y$

So:

$
\frac{x+y}{y}=\left[\frac{16}{12}\right]^u=\left[\frac{4}{3}\right]^u
$

and:

$
\frac{y}{x}=\left[\frac{12}{9}\right]^u=\left[\frac{4}{3}\right]^u
$

Therefore if $z=x/y$, we have:

$
z+1=1/z
$

which is a quadratic in $z$, one (or more of which's solutions)
should solve the problem.

RonL

RonL