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.
$\displaystyle log\cdot x^{\frac{1}{2}} - log \cdot x^{\frac{1}{3}} = log\,2$
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.
$\displaystyle log_{4}\,(x + 2) + log_{4}\,(x - 3) = log_{4}\,9$
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: $\displaystyle (x + 2)\,(x - 4) = 9 \Rightarrow x^2 -2x - 17 = 0$
Use the quadratic formula.
I got $\displaystyle x =\pm 3\sqrt{2} + 1$
Check to see if both solutions work.