1. ## Log

I need without calculator improve that: $\frac{1}{log_35}+\frac{1}{log_95}>2$

what I need to do with $log_?5$ ?

2. I would try to write both logarithms using the same base. It's very awkward using different bases. How could you do that?

3. Use this formula: $log_am=\frac{logm}{loga}$?
so, $\frac{log5}{log3}$ and $\frac{1}{2}\frac{log5}{log3}$

it's this or not?

4. You're correct, but you can go further. I would agree that

$\log_{9}(5)=\dfrac{\ln(5)}{\ln(9)}=\dfrac{\ln(5)}{ \ln(3^{2})}=\dfrac{\ln(5)}{2\ln(3)}=\dfrac{1}{2}\l og_{3}(5).$

So now what can you do?

5. Originally Posted by Ackbeet
You're correct, but you can go further. I would agree that

$\log_{9}(5)=\dfrac{\ln(5)}{\ln(9)}=\dfrac{\ln(5)}{ \ln(3^{2})}=\dfrac{\ln(5)}{2\ln(3)}=\dfrac{1}{2}\l og_{3}(5).$
with this $=log_3\sqrt{5}$

6. You can do that, but I'm not sure I'd recommend it. The reason is that if you leave it at $\dfrac{1}{2}\log_{3}(5),$ then it's easier to get a common denominator on the LHS of the original expression.

So, take a look at the original expression, and substitute in this $\dfrac{1}{2}\log_{3}(5).$ What do you get?

7. So, I have this $\frac{1}{log\frac{5}{3}}+\frac{1}{\frac{1}{2}log_3 5}>2$

8. Not quite. I would write this:

$\dfrac{1}{\log_{3}(5)}+\dfrac{1}{\dfrac{1}{2}\log_ {3}(5)}>2.$

(No need to change the first fraction.) So that is what we need to prove. What would you do next?

9. Can I use: $\frac{1}{log_ma}=log_am$
then $log_53+\frac{1}{2}log_53>2$
or
$log_53+log_59>log_525$
$log_527>log_525$
$27>25$

10. Sure you can. However, you would get

$\log_{5}(3)+2\log_{5}(3)>2,$ not what you got.