Manipulating log function I'm trying to find the value

Axel Stout

Axel Stout

Answered question

2022-04-13

Manipulating log function
I'm trying to find the value of a constant for y(0)=0 in the following differential equation.
2ln(2x+3y1)2x+3y2=2x+3y+k
Of course when plugging in the values, I get 2ln(1)=k which errors. When entering this into Wolfram Alpha, they suggest rearranging the equation from this format,
2ln(x1)x2+constant
to
x2+2ln(1x)+12+constant
"Which is the equivalent for restricted x values", which indeed I have. This would leave me with a positive ln(1), which would solve my problem... but...
My question, how does this manipulation work? I've never seen this before. I don't understand how they've made that leap (or if it's even accurate).
Can anyone educate me?

Answer & Explanation

Assorrymarf0cgr

Assorrymarf0cgr

Beginner2022-04-14Added 10 answers

Note that for any constants a,b,cR
aln(b(x+c))=aln(b)+aln(x+c)=ln(ba)+ln((x+c)a)
Therefore, in this case we have:
2ln(x1)x2+c=2ln(1(1x))x2+c=ln((1)2(1x)2)x2+c
And using the fact that (1)2=1 and ln(1)=0 we have:
2ln(x1)x2+c=2ln(1x)x2+c
And then as amWhy points out, you can have c=12+k for restricted values of x.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?