1. ## Logarithm question

If X and Y are both positive and log9X=log12Y=log16(X+Y), find the value of X/Y.

2. ## Re: logarithm problem

9, 12 and 16 are base numbers or not?

3. Originally Posted by yobacul
9, 12 and 16 are base numbers or not?
Yes.

I am trying using the [tex] equation below, hopeful it works.

If x and y are both positive and $\displaystyle \log_9 x = \log_{12} y = \log_{16} (x+y)$ find the value of x/y.

4. $\displaystyle \log_9 x = \log_{12} y = \log_{16} (x+y)$
or

$\displaystyle \frac{log_{10} x}{log_{10} 9}=\frac{log_{10} y}{log_{10} 12}=\frac{log_{10} (x+y)}{log_{10} 16}$
on solving

$\displaystyle \frac{log_{10} x}{log_{10} 9}=\frac{log_{10} (x+y)}{log_{10} 16}$

$\displaystyle \frac {log_{10} 16}{log_{10} 9}=\frac{log_{10} (x+y)}{log_{10} x}$

i stuck here

5. Thanks, but I think there is an error as shown below:

$\displaystyle \frac {log_{10} 16}{log_{10} 9}$

is not equal to $\displaystyle log_{10}(16-9)$

6. Originally Posted by yeoky
Thanks, but I think there is an error as shown below:

$\displaystyle \frac {log_{10} 16}{log_{10} 9}$

is not equal to $\displaystyle log_{10}(16-9)$
.

7. Originally Posted by yeoky
Thanks, but I think there is an error as shown below:

$\displaystyle \frac {log_{10} 16}{log_{10} 9}$

is not equal to $\displaystyle log_{10}(16-9)$
It most certainly is NOT. If it were true then it would also be true that $\displaystyle \frac{\log_{10} 100}{\log_{10} 10} = \log_{10}(100 - 10)$ ....!!