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

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)]

stapel_eliz
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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...?

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"...?

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

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

stapel_eliz
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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?

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

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 $\log_x(ax+b)=2$ you could still solve