# Logarithms question

1. Dec 11, 2004

### 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.

Last edited: Dec 11, 2004
2. Dec 11, 2004

### 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

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

3. Dec 11, 2004

### 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

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4. Dec 11, 2004

### 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)?

5. Dec 11, 2004

### seiferseph

now i do its supposed to be log_5(36)