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Solution 3.4:2c

From Förberedande kurs i matematik 1

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Regardless of what value x has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get x down from the exponents,

LHSRHS=ln3ex2=ln3+lnex2=ln3+x2lne=ln3+x2,=ln2x=xln2.

If we collect the terms onto one side, the equation becomes

x2xln2+ln3=0

which is a standard second-order equation for which we complete the square

x21ln2221ln22+ln3=0,x21ln22=21ln22ln3.

Now, we have to be cautious and remember that, because 2e3 and thus ln21ln3 we have that 41(ln2)2ln3 and the right-hand side is therefore negative. Since the square on the left-hand side cannot be negative, the equation has no solution.