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ln(x-1)+ln(x+2)=1
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I think the first step is to use the ln properties: ln(x-1)(x+2)=1
what you have to do here? solve for x?
yes
Then you can undo the natural log by putting them as powers of e. (x-1)(x-2)=e Then solve as you would any other equation.
x=1+e and x=e-2
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Corrections to my solutions. x is not equal to 1+e or e-2. (x-1)(x-2)-e=0 I have to use the quadratic formula to solve. x=1.728964295 and x=-2.728964295 Note the negative solution is extraneous (not valid).
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