a-level maths, logarithms

e^(2y) = x + 1
ln(x-2) = 2y - 1

does anyone know how to solve these simultaneous equations

what have you tried so far ?

i have cried

This is a mood.

Notice how there is something in common with both equations, the trick is to remove the exponential - do you remember how to do this?

no
we just started this topic during lockdown so i've had to teach it to myself and that has not gone well

Okay so the first thing to understand is that exponentials and logarithms are linked but can undo each other (in a sense).

If you have y=e^x and want to get the x, take the ln of each side and you get ln(y) = x. This is because if you take the natural log of an exponential you will get the exponent (i.e. the natural log will undo the exponential).

ok so would i do 2y = In(x+1), y = (In(x+1))/2
then substitute???
ln(x-2) = 2y - 1
ln(x-2) = In(x+1) - 1
i don't know where to go from there

Perfect! Now you need to use one of the log laws, the one relevant here is: ln(a) - ln(b) = ln(a/b)

Then you will have all your Xs in a single ln. So to get rid of a natural log, for example ln(a), simply take the exponential of it: i.e. e^ln(a) = a

ohh!! i got it now i think
x = (e+2)/(1-e)
and then i just substitute into e^(2y) = x + 1
e^(2y) = (e+2)/(1-e) + 1
then im stuck

If you have ln(x-2) = In(x+1) - 1:

1) rearrange it to make it more friendly
ln(x-2) = In(x+1) - 1
=> 1 = ln(x+1) - ln(x-2)

2) use the log law: ln(a) - ln(b) = ln(a/b)

this should give you something like 1 = ln(q) where q is determined by using the log law previously

3) then take the exponential of the log: the general rule is if x = ln(y) then e^x = y

for your question you will end up with e^1 = q, at this point it should be easy to just rearrange to find x.

oh wait yeah sorry
so x = (1 +2e)/(e - 1) maybe
and then substitute into e^(2y) = x+1
y = 0.5 * In((1+2e)/(e-1) +1)
is that good?

That's what I got for x, so yep I think so! You can probably simplify your answer a bit but unless the question asks you to do so I wouldn't bother.