# Logs two equtions

• Apr 24th 2010, 04:54 PM
mathwhat
Logs two equtions
$\displaystyle \log_3(x)+\log_9(y)=1$
$\displaystyle \log_x(3)+\log_y(9)=-\frac{1}{2}$

How can i solve this? (Thinking)

Thank you!
• Apr 24th 2010, 05:13 PM
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Quote:

Originally Posted by mathwhat
$\displaystyle \log_3(x)+\log_9(y)=1$
$\displaystyle \log_x(3)+\log_y(9)=-\frac{1}{2}$

How can i solve this? (Thinking)

Thank you!

Are you asking to solve these equations simultaneously for $\displaystyle x$ and $\displaystyle y$?

I'd try converting everything to the natural logarithm using the change of base rule $\displaystyle \log_b{x} = \frac{\log_k{x}}{\log_k{b}}$.

So Equation 1:

$\displaystyle \log_3{x}+\log_9{y}=1$

$\displaystyle \frac{\ln{x}}{\ln{3}} + \frac{\ln{y}}{\ln{9}} = 1$

$\displaystyle \frac{\ln{x}}{\ln{3}} + \frac{\ln{y}}{2\ln{3}} = 1$

$\displaystyle \frac{2\ln{x} + \ln{y}}{2\ln{3}} = 1$

$\displaystyle 2\ln{x} + \ln{y} = 2\ln{3}$

$\displaystyle \ln{y} = 2\ln{3} - 2\ln{x}$.

Equation 2:

$\displaystyle \log_x{3}+\log_y{9}=-\frac{1}{2}$

$\displaystyle \frac{\ln{3}}{\ln{x}} + \frac{\ln{9}}{\ln{y}} = -\frac{1}{2}$

$\displaystyle \frac{\ln{3}}{\ln{x}} + \frac{2\ln{3}}{\ln{y}} = -\frac{1}{2}$

$\displaystyle \frac{\ln{3}\ln{y} + 2\ln{3}\ln{x}}{\ln{x}\ln{y}} = -\frac{1}{2}$

$\displaystyle 2\ln{3}\ln{y} + 4\ln{3}\ln{x} = -\ln{x}\ln{y}$

$\displaystyle 2\ln{3}\ln{y} - \ln{x}\ln{y} = -4\ln{3}\ln{x}$

$\displaystyle \ln{y}(2\ln{3} - \ln{x}) = -4\ln{3}\ln{x}$

$\displaystyle \ln{y} = \frac{-4\ln{3}\ln{x}}{2\ln{3} - \ln{x}}$.

Now you can set the equations equal to each other and solve.
• Apr 24th 2010, 08:03 PM
mathwhat
Thank you but i don't familiar with $\displaystyle \ln$...
I am only know the $\displaystyle \log$ thing...
• Apr 24th 2010, 08:06 PM
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Quote:

Originally Posted by mathwhat
Thank you but i don't familiar with $\displaystyle \ln$...
I am only know the $\displaystyle \log$ thing...

$\displaystyle \ln$ is just shorthand for the logarithm with base $\displaystyle e$.

I.e. $\displaystyle \ln{x} = \log_e{x}$.

If you want you can use the base 10... It doesn't make any difference. The point of using the change of base rule is to get all the logarithms to HAVE THE SAME BASE.
• Apr 25th 2010, 07:52 AM
mathwhat
Thank you very much for the useful information!!! (Happy)
• Apr 29th 2010, 06:59 AM
icefirez
hi this is my answer i dont know its right cuz im not good at math...but hopefully it's right from my mention
just anohter thing i dont know to much latex language so ....

$\displaystyle log_3{x}+ log_9{y} = 1$
$\displaystyle log_3{x}+log_3^2{y}=1$
here is log base 3 SQUARED of y but cant do it with latex
$\displaystyle \log_3{9}=\log_3{3^2}=2\log_3{3}=2$
its another role of changing base now example
$\displaystyle \log_9{3}=\log_3^2{3}=\frac{1}{2}\log_3{3}=\frac{1 }{2}$

$\displaystyle log_3{x}+\frac{1}{2}log_3{y}=log_3{3}$

we know that log base 3 of number 3 is 1 so we can write 1 as log base 3 of 3

now we remove logs and we got

$\displaystyle x+\frac{1}{2}y=1$
hope im right and you solve other equation