I'm having problems with these equations in particular:

(logx^1/2)-(logx^1/3)=log2

and

log(base4)(x+2)+log(base4)(x-3)=log(base4)9

Thanks.

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- May 25th 2008, 05:21 PMandymacSolving Log Equations
I'm having problems with these equations in particular:

(logx^1/2)-(logx^1/3)=log2

and

log(base4)(x+2)+log(base4)(x-3)=log(base4)9

Thanks. - May 25th 2008, 05:39 PMJonboy

Focus on the left of the equal sign. When you subtract logs, and they have like bases, you may divide what you are taking the log of.

So:

Now the logs cancel, and you're left with:

Now cube both sides:

That simplifies to:

Square both sides, and so is 64.

Plug it back in to make sure it's correct. - May 25th 2008, 05:44 PMandymac
Thanks a lot. That helps. Could you show me how to do the second one too? I think it involves factoring and I couldn't figure it out.

- May 25th 2008, 05:48 PMJonboy

Once again, focus on the left side of the equal sign. When you add logs with like bases, you may multiply what you are taking the log of.

So:

Logs cancel:

Use the quadratic formula.

I got

Check to see if both solutions work. - May 25th 2008, 05:49 PMJonboy
and when i say "plug it back in..." i mean in the original problem. make sure you have no extraneous, or false, solutions.

- May 25th 2008, 05:53 PMandymac
So: http://www.mathhelpforum.com/math-he...8ef2f594-1.gif

Logs cancel: http://www.mathhelpforum.com/math-he...5294fef5-1.gif

How did you go from log(base4)(x+2)(x-4) to (x+2)(x-4)? shouldnt it still be (x+2)(x-4)?

If you the part that confuses me is if you have (x+2)(x-4) because then you have an x factor to deal with also. - May 25th 2008, 05:54 PMandymac
or wait sorry, originally it was (x+2)(x-3)

- May 25th 2008, 06:03 PMJonboy
i messed up. got too wishful. ;)

i edited my post. i think it's correct now. - May 25th 2008, 06:06 PMandymac
Thanks So Much