Lösung 3.4:3c

Aus Online Mathematik Brückenkurs 1

With the log laws, we can write the left-hand side as one logarithmic expression,

lnx+lnx+4=lnxx+4 

but this rewriting presupposes that the expressions ln x and lnx+4  are defined, i.e. x0 and x+40. Therefore, if we choose to continue with the equation

lnxx+4=ln2x+3 

we must remember to permit only solutions that satisfy x0 (the condition x+40 is then automatically satisfied).

The equation rewritten in this way is, in turn, only satisfied if the arguments xx+4   and 2x+3 are equal to each other and positive, i.e.

xx+4=2x+3 

We rewrite this equation as x2−2x−3=0 and completing the square gives

x+12−12−3=0x+12=4

which means that x=−12, i.e. x=−3 and x=1.

Because x=−3 is negative, we neglect it, whilst for x=1 we have both that x0 and xx+4=2x+30 . Therefore, the answer is x=1.