logarithm questions

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'm also confused about

log_2(y-3x+2)=0

I changed it to

\frac{ln(y-3x+2)}{ln2}=0

would appreciate any help!

(edited 9 years ago)

Reply 1

Raysorb

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

Reply 2

tombayes

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'm also confused about

log_2(y-3x+2)=0

I changed it to

\frac{ln(y-3x+2)}{ln2}=0

would appreciate any help!

whats the question?
make

x

the subject of

2^x - 3^y = -1

Reply 3

tombayes

Original post by Raysorb

Hi!

when you take log of both sides you get
xlog 2 - ylog3= -log 1

no you don't :frown:

Reply 4

HelloGoodbye

Original post by tombayes

whats the question?
make

x

the subject of

2^x - 3^y = -1

the question is solve the following simultaneous equations:

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

(edited 9 years ago)

Reply 5

davros

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've already posted this question once in another thread!

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

log(a - b) \neq log a - log b

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.