# Thread: Logarithm - please help

1. ## Logarithm - please help

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.

2. Originally Posted by madaboutmath
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:

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

Then:

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

So:

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

and:

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

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

$\displaystyle z+1=1/z$

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

RonL

RonL

3. ## Logarithm

I really appreciate your prompt help CaptainBlack. The answer is (sqrt5-1)/2.

Thanks.

MAM