logs is driving me crazy! solve log3(5x-3)-log3(x+3)=1

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
MissT
Posts: 4
Joined: Fri Apr 17, 2009 4:32 pm
Contact:

logs is driving me crazy! solve log3(5x-3)-log3(x+3)=1

Postby MissT » Fri Apr 17, 2009 7:49 pm

How do i solve the following:
log3(5x-3)-log3(x+3)=1

and

logx(9x+10)=2

and

[(256^1/4)+(log5 625)+(2x^0)]

User avatar
stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Fri Apr 17, 2009 8:15 pm

MissT wrote:1) log3(5x-3)-log3(x+3)=1

The first step will be to combine the subtracted logs into one log containing a division, by applying the appropriate log rule.

Once you have "log, base three, of (a rational expression)", convert this to the equivalent exponential form, using the definition of logs: "three, to the power one, equals (a rational expression)".

Then solve the resulting rational equation.

MissT wrote:2) logx(9x+10)=2

I'm afraid I don't know what they want you to do with a "base x" log...? :confused:

MissT wrote:3) [(256^1/4)+(log5 625)+(2x^0)]

This one cannot be "solved", because it isn't an equation: this is no "equals" sign in this. Are you perhaps supposed to "evaluate"...? :oops:

MissT
Posts: 4
Joined: Fri Apr 17, 2009 4:32 pm
Contact:

Re: logs is driving me crazy! solve log3(5x-3)-log3(x+3)=1

Postby MissT » Fri Apr 17, 2009 8:24 pm

i'm suppose to solve for x in all of them

User avatar
stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Fri Apr 17, 2009 11:46 pm

As mentioned previously, it is not possible to "solve" the third exercise, since it is not an equation. So please get corrected instructions from your teacher.

As for the first two, you've been given the steps and been provided with lessons for the review of any bits on which you may be "fuzzy". What progress have you made?

Please be complete. Thank you! :popcorn:

User avatar
Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

Re: logs is driving me crazy! solve log3(5x-3)-log3(x+3)=1

Postby Martingale » Sat Apr 18, 2009 12:44 am

MissT wrote:How do i solve the following:
log3(5x-3)-log3(x+3)=1

and

logx(9x+10)=2


I'll give you a little hint...the solution to both of these are positive integers. the first one should be easy to solve for and the second is a little trickier...but you can still solve for it. even if it was you could still solve :D


Return to “Intermediate Algebra”