How do I solve log(base 5)36 in terms of x and y?

[seiferseph]

if log2 = x and log3 = y, solve for log(base5)36 in terms of x and y.

how do i even get started on this? I'm really confused with logs.

 

[Tide]

This seems like a rather roundabout way of doing things. In any case ...

I assume those logarithms are base 10 in which case, by definition of logarithm, we have

$$10^x = 2$$
$$10^y = 3$$

Let $z = \log_5 10$ which means $5^z = 10$. But

$$5 = 2 + 3 = 10^x + 10^y$$

so that

$$5^z = \left(10^x + 10^y\right)^z = 10$$

IOW I'm not sure I see the point of the problem!

from which

$$z \log \left( 10^x + 10^z\right) = \log 10 = 1$$

and finally

$$z = \frac {1}{\log \left( 10^x + 10^z \right)}$$


Of course, a direct approach would have led to

$$\log_5 10 = \frac {\log 10}{\log_{10} 5} = \frac {1}{\log 5}$$

 

[seiferseph]

thanks, but I'm not sure if that is correct (the teacher actually said it was quite simple). the last question was something like a^2 + b^2 for the answer. i'll post a little bit of what i got, I'm not sure if this is right

 

[Muzza]

Do you realize that your original question is about log_5(10), but the handwritten thing you posted now is about log_5(36)?

 

[seiferseph]

Do you realize that your original question is about log_5(10), but the handwritten thing you posted now is about log_5(36)?

now i do its supposed to be log_5(36)