Question

Review: Solving Logarithmic Functions
Question below.

Answer 1

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?

Answer 2

causes you can't take the of a negative number

Answer 3

oops should be \(\sf ln\ e^{2x}\)

oh yeah i got it lol thanks

Answer 4

i tried it once, had a headache for three diays

Answer 5

yw