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# logarithm questions Watch

1. I'm trying to do these questions but I have no idea how to:

2^x - 3^y = -1
I changed it to xln2 - yln3 = -1 and I don't know what to do next :/

I changed it to

would appreciate any help!
2. Hi!
Our course might not be the same but I have some idea about logarithms(not in my course tho)
when you take log of both sides you get
xlog 2 - ylog3= -log 1

now log 1 =0, therefore xlog 2 =y log 3
or x/y = log 3/ log 2
3. (Original post by HelloGoodbye)
I'm trying to do these questions but I have no idea how to:

2^x - 3^y = -1
I changed it to xln2 - yln3 = -1 and I don't know what to do next :/

I changed it to

would appreciate any help!
whats the question?
make the subject of
4. (Original post by Raysorb)
Hi!

when you take log of both sides you get
xlog 2 - ylog3= -log 1
no you don't
5. (Original post by tombayes)
whats the question?
make the subject of
the question is solve the following simultaneous equations:

2^x - 3^y = -1
2^(x-1) - 3^(y-1) = -1
6. (Original post by HelloGoodbye)
the question is solve the following simultaneous equations:

2^x - 3^y = -1
2^(x-1) - 3^(y-1) = -1

You can't just take logs the way you think you can -

and log(-1) is undefined (at least in school maths).

Rearrange the first eq to get 2^x in terms of 3^y (or vice versa) and substitute into the 2nd eq.

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