Processing Math: 76%
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.

jsMath

Solution 3.4:3c

From Förberedande kurs i matematik 1

Revision as of 14:34, 2 October 2008 by Tek (Talk | contribs)
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
Jump to: navigation, search

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

lnx+ln(x+4)=ln(x(x+4))

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

ln(x(x+4))=ln(2x+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 x(x+4) and 2x+3 are equal to each other and positive, i.e.

x(x+4)=2x+3.

We rewrite this equation as x2+2x3=0 and completing the square gives

(x+1)2123(x+1)2=4=0

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

Because \displaystyle x=-3 is negative, we neglect it, whilst for \displaystyle x=1 we have both that \displaystyle x > 0 and \displaystyle x(x+4) = 2x+3 > 0\,. Therefore, the answer is \displaystyle x=1\,.