precalc: loga(x-1)-loga(x+6)=loga(x-2)-loga(x+3)

[hummingbyrd88]

ok so I'm new here. this is my second semester taking precalculus so you can probably see i need all the help i can get. here is my question/problem:

loga(x-1)-loga(x+6)=loga(x-2)-loga(x+3)

now, what i thought i was supposed to do was to first combine the logarithmic equations and make both sides one big equation:

loga (x-1)/(x+6)= loga (x-2)/(x+3)

but after i do that, i don't know where to go from there.

 

[Loren]

Re: another precalculus question

If log a = log b, doesn't a=b?

 

[soroban]

Hello, hummingbyrd88!

Welcome aboard!

$\displaystyle \log_a(x-1)-\log_a(x+6) \;=\;\log_a(x-2)-\log_a(x+3)$

Here's a step that saves us a lot of grief . . . Rearrange the terms.


$\displaystyle \text{We have: }\;\log_a(x-1) + \log_a(x+3) \;=\;\log_a(x-2) + \log_a(x+6)$

$\displaystyle \text{Then: }\;\log_a\bigg[(x-1)(x+3)\bigg] \;=\;\log_a\bigg[(x-2)(x+6)\bigg]$

. . See? We avoid those fractions and quotients.


$\displaystyle \text{As Loren suggested, we can 'un-log' both sides: }\;(x-3)(x+3) \;=\;(x-2)(x+6)$


Go for it!