Thread: Solving for x in a logarithm

log(base3)x + log(base3)(2x+51) = 4

a) thank you for helping, I am a new member and love the looks of this site.
b) i am not asking you to do this, maybe just some help I am stuck
c) its not really homework, i am studying for a test on monday and just cant understand this

1) I know the rules of logs and this is where i get stuck:

log(base3)(2x+51)(x)=4
log(base3)2x^2+51x=4
2x^2+51x=3^4 --->2x^2+51x=81
? now do i divide both sides by 2? or should I have done that before i got rid of log(base3)?

p.s. I know how to factor, but not something like 2x^2+51x-81=0

Any help would be amazing, even a little input if your not sure to get me thinking. Thanks!

p.s. I know how to factor, but not something like 2x^2+51x-81=0

2x^2 + 51x - 81 = 0

(2x - 3)(x + 27)

Wow, always getting caught up on the simple steps.
Thanks for helping me out. When its something like that I always get thrown off a bit.

Originally Posted by dpjwilson

2x^2+51x=3^4 --->2x^2+51x=81

p.s. I know how to factor, but not something like 2x^2+51x-81=0

Before your test, you might want to review some online lessons to refresh your memory regarding factoring quadratics.

Good luck on the test!