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)]

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)]

log3(5x-3)-log3(x+3)=1

and

logx(9x+10)=2

and

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

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

The first step will be to combine the subtracted logs into one log containing a division, by applying the appropriate1) log3(5x-3)-log3(x+3)=1

Once you have "log, base three, of (a rational expression)", convert this to the equivalent

Then

I'm afraid I don't know what they want you to do with a "base x" log...?2) logx(9x+10)=2

This one cannot be "solved", because it isn't an equation: this is no "equals" sign in this. Are you perhaps supposed to "evaluate"...?3) [(256^1/4)+(log5 625)+(2x^0)]

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

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

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!

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!

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
**Contact:**

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 solveHow do i solve the following:

log3(5x-3)-log3(x+3)=1

and

logx(9x+10)=2