Logs

[Jonathan B]

All logs here are to base 3. how do i solve

2Logy-Log(y+4)-2=0

Any help appreciated

[Mattywoo2]

2 log(y) - log(y+4) = log(y²/y+4)

So our equation becomes

log(y²/y+4) = 2

y²/y+4 = 3²

y² = 9y + 36

This is a simple quadratic in y.

[eranga262154]

Dear Jonathan B,

Wellcome to the forum. Those things you are request here is the basis of logrithms. Try to learn those first.

[triggernum5]

I think he's stuck with the fact that vb only has a builtin log10() function.. If thtas the case, Log3(x) = Log10(x)/Log10(3)

[eranga262154]

I think he's stuck with the fact that vb only has a builtin log10() function.. If thtas the case, Log3(x) = Log10(x)/Log10(3)

Ya, it's true. But in the fundamentals I think anyone not teach it using log10, log3, etc: Just used a comman letter for it. It is the best way.

[triggernum5]

Actually, I find that ppl weak in math tend to learn better with fewer variables and more pure examples.. But ok here goes...
Log_b(x) = Log_a(x)/Log_a(b)

[eranga262154]

Correct