Please help

G

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Guest
My name is Amanda Lynn. I am new to this. My teacher told me to try it out and ask this question because I keep missing it on my homework. She said someone could show me how to do it correctly. Thanks.

If ln(x - 11) - ln(x - 3) = ln (x + 2) - ln(x + 12), then x =
 
First lets rewrite the equation.
ln(x−11)+ln(x+12)=ln(x+2)+ln(x−3)
Using the fact that ln(ab)=ln(a)+ln(b) we get:
ln[(x−11)(x+12)]=ln[(x+2)(x−3)].
Now on its domain the ln() function is one-to-one, the domain here is x>11.
So we get [(x−11)(x+12)]=[(x+2)(x−3)] or x<SUP>2</SUP>+x−132= x<SUP>2</SUP>−x−6.
Now you solve for x.
 
Hello, AmandaLynn04!

Welcome aboard!

Solve for \(\displaystyle x:\;\;\ln(x - 11)\,-\,\ln(x - 3) \:= \:\ln(x + 2)\,-\,\ln(x + 12)\)
Click to expand...
I would rearrange terms first *: .\(\displaystyle \ln(x\,-\,11)\,+\,\ln(x\,+\,12) \;= \;\ln(x\,-\,3)\,+\,\ln(x\,+\,2)\)

Then we have: .\(\displaystyle \ln[(x\,-\,11)(x\,+\,12)] \:= \:\ln[(x\,-\,3)(x\,+\,2)]\)

. . . . . . . . . . . . .or: .\(\displaystyle \ln(x^2\,+\,x\,-\,132) \:= \:\ln(x^2\,-\,x\,-\,6)\)

Take anti-logs: . . . . . . .\(\displaystyle x^2\,+\,x\,-\,132 \:= \:x^2\,-\,x\,-\,6\)

. . . Solve for \(\displaystyle x:\;\;2x\,=\,126\qquad\Rightarrow\qquad x\,=\,63\)

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*
I moved some of the terms around . . . why?

I moved the minus-terms to the other side of the equation.
. . That way, I avoided the divisions . . . see?

[I bet your teacher doesn't know enough to do that.
It takes years of dedicated Laziness to come up with that, you see.]

(Edit: too fast for me, pka!)
 
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