1. ## 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.

2. $\displaystyle log\cdot x^{\frac{1}{2}} - log \cdot x^{\frac{1}{3}} = log\,2$

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

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

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

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

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

Plug it back in to make sure it's correct.

3. 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.

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

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

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

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

Check to see if both solutions work.

5. and when i say "plug it back in..." i mean in the original problem. make sure you have no extraneous, or false, solutions.

6. So:

Logs cancel:

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.

7. or wait sorry, originally it was (x+2)(x-3)

8. i messed up. got too wishful.

i edited my post. i think it's correct now.

9. Thanks So Much