Question

log5(x)+log3(x)=1

Answer 1

what is log m+log n=?

Answer 2

they are multiplied

Answer 3

but what to do with log base 5 and log base 3 since they are different?

Answer 4

so the question is

\[\log_{5}(x) + \log_{3}(x) = 1\]

Answer 5

Yes campbell_st

Answer 6

ok... you need to use change of base.... and change them to either base 10 or base e

the change of base formula is

\[\log_{a}(m) = \frac{\ln(m)}{\ln(a)}\]

and ln(a) becomes just a number....

hope it helps

Answer 7

log m + log n = log mn is what @kaylady21 meant.

Answer 8

Yes, but they have different bases.

Answer 9

Good point indeed!! The bases ARE different. Therefore, we must choose one base to keep and must write the other log in terms of the one we keep.

Answer 10

How do you do that? :)

Answer 11

Before we do any more, let me verify that the bases here are 3 and 5. Is that what you meant?

Answer 12

Yes, that is what I meant @mathmale

Answer 13

Then follow what @campbell_st said earlier.

Answer 14

\[\log_{5} x+\log_{3}x=1 \]

Answer 15

If that's correct, let's stick with one base: 3.

We must then convert that base 5 log to base 3. @klady21, can you convert (log to the base 5 of x to base 3?

Answer 16

Or better yet, apply this following rule: \[\log_{b}a =\frac{ 1 }{ \log_{a}b } \]

Answer 17

Example, \[\log_{5}x =\frac{ 1 }{ \log_{x} 5 } \]

Answer 18

As long as x > 0; x ≠ 1

Answer 19

@calculusfunctions: would you mind demonstrating how you'd convert log-to-the-base-3 into a base e expression?

I'm used to the change of base formula that reads like this:\[\log_{a} x=\frac{ \ln x }{ \ln a }\]

Answer 20

And thus, I'd convert \[\log_{5} x\] as follows:\[\log_{5}x=\frac{ \log_{3} x }{ \log_{3} 5 } \]

Answer 21

So the expression \[\log_{5}x+\log_{3}x=1 \]

Answer 22

Alright then, \[\log_{3}x =\frac{ \ln x }{ \ln 3 } \]

Answer 23

would become\[\frac{ \log_{3} x }{ \log_{3} 5 }+\log_{3}x=1 \]

Answer 24

@calculusfunctions : Are we in agreement with this last result?

Answer 25

Absolutely @mathmale You are correct!

Answer 26

OK, then obviously our next step is to solve for log-to-the-base-3-of-x, and, finally, for x itself.

@kaylady21? What would you suggest we do next? If you're sure of what to do, do it, please.

Answer 27

Remember @mathmale and everyone else, that the important thing is to lead her to the correct solution. NOT DO IT for her!!!!

Answer 28

@reemii: I have trouble with the fact that you've put log x into the denominators of two fractions.

Answer 29

@reemii you are also correct. It's the same as what @mathmale said.

Answer 30

There has to be a mathematically logical reason for every step we take.

Answer 31

@kaylady21: we have to be careful not to take your math problem and run with it. Could you let the rest of us know what you are thinking at this point?

Answer 32

indeed.
x=1 is forbidden at the moment.
but we can see in the first equation that x=1 is not a solution. so we can go on safely.

however this must be said. thanks @mathmale

Answer 33

Yes @mathmale I just realized that. The log x is not in the denominators.

Answer 34

I am getting confused. I got to the point you did but after that....I'm not sure.

Answer 35

Do you need to find like denominators?

Answer 36

Not really. Let me go back to what I proposed before:

Answer 37

\[\frac{ \log_{3}x }{ \log_{3}5 }+\log_{3} x=1\]

Answer 38

Note that both of the terms on the left have log=-to-the-base-3-of-x in them. This suggests that we could factor that common term out. Anyone care to try doing that?

Answer 39

Remember, our goals are first to find log-to-the-base-3 of x first and x itself second.