# Solving Log Equations

• May 25th 2008, 05:21 PM
andymac
Solving 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 PM
Jonboy
$log\cdot x^{\frac{1}{2}} - log \cdot x^{\frac{1}{3}} = log\,2$

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: $log\,\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} = log\,2$

Now the logs cancel, and you're left with: $\frac{\sqrt{x}}{\sqrt[3]{x}} = 2$

Now cube both sides: $\frac{\sqrt{x}\cdot\sqrt{x}\cdot \sqrt{x}}{x} = \frac{x\cdot \sqrt{x}}{x} = 8$

That simplifies to: $\sqrt{x} = 8$

Square both sides, and so $x$ is 64.

Plug it back in to make sure it's correct.
• May 25th 2008, 05:44 PM
andymac
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 PM
Jonboy
$log_{4}\,(x + 2) + log_{4}\,(x - 3) = log_{4}\,9$

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: $log_{4}\,(x + 2)(x-3) = log_{4}\,9$

Logs cancel: $(x + 2)\,(x - 4) = 9 \Rightarrow x^2 -2x - 17 = 0$

I got $x =\pm 3\sqrt{2} + 1$

Check to see if both solutions work.
• May 25th 2008, 05:49 PM
Jonboy
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 PM
andymac
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 PM
andymac
or wait sorry, originally it was (x+2)(x-3)
• May 25th 2008, 06:03 PM
Jonboy
i messed up. got too wishful. ;)

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