help logs

[tawdry moat]

For this, I tried to divide it but then where does the 2 infront of the log go??

[fast acorn]

I would use the power rule to bring the 2 up

so log_2 x^2 - log_2 (x+3) = 2

then use subtraction law

[tawdry moat]

But then I got log_2(x/3) = 2

[fast acorn]

you should have log_ 2 x²/(x+3) = 2

[tawdry moat]

Yeah exactly

[fast acorn]

now how do we undo a log

[tawdry moat]

Cancel it out on both sides

So the 2 needs a log

[fast acorn]

I don't think so

[tawdry moat]

Huh

[fast acorn]

logarithms are inverse of exponentials

so we can do the base of the log to the power of the rhs integer

so x²/(x+3) = 4

[tawdry moat]

I did this

[fast acorn]

you can't cancel the x's

[tawdry moat]

Why not

[fast acorn]

algebra does not work like that unfortunately

[tawdry moat]

That’s strange

[fast acorn]

because the expression is on the denominator

[tawdry moat]

Yeah

[fast acorn]

you could split the fraction if the x² was the denominator

but you can't cancel with two terms or more on the denominator

just doesn't work

[tawdry moat]

I’ve done this but now what’s wrong

[fast acorn]

nothing

[tawdry moat]

Oh alright

[fast acorn]

but think about your answers

[tawdry moat]

It’s just that the website says it’s wrong

[fast acorn]

one of them is the answer

can you have a negative log

[tawdry moat]

Oh you cant

[fast acorn]

exactly

[tawdry moat]

So just 6

[fast acorn]

so there's 1 solution, x = 6

precisely

[tawdry moat]

Ok thanks

[fast acorn]

np