## anonymous one year ago Review: Solving Logarithmic Functions Question below.

1. anonymous

Help me solve this please. $$\sf \sqrt{ln(e^2x)+12}-2=e^{ln x}$$ I know that $$\sf ln e^(2x)=2x$$ and $$\sf e^{ln x}=x$$ So $$\sf \sqrt{2x+12} -2=x \\ \sqrt{2x+12}=x+2$$ then I square on both sides to get rid off the radical: $$\sf 2x+12=(x+2)^2 \\ 2x+12=x^2+4x+4\\ 0=x^2+4x-2x+4-12 \\ 0=x^2+2x-8 \\ (x+4)(x-2)$$ so x=-4 and x=2.. but how come the correct answer is only x=2? Did I do something wrong?

2. anonymous

causes you can't take the of a negative number

3. anonymous

oops should be $$\sf ln\ e^{2x}$$ oh yeah i got it lol thanks

4. anonymous