Processing Math: Done
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, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | \text{LHS} | ||
| + | &= \ln\bigl( 3e^{x^{2}}\bigr) | ||
| + | = \ln 3 + \ln e^{x^2} | ||
| + | = \ln 3 + x^2\ln e | ||
| + | = \ln 3 + x^2\,\textrm{,}\\[5pt] | ||
| + | \text{RHS} | ||
| + | &= \ln 2^x = x\ln 2\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | If we collect the terms onto one side, the equation becomes | ||
| + | |||
| + | {{Displayed math||<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}} | ||
| + | |||
| + | which is a standard second-order equation for which we complete the square | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt] | ||
| + | \Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | Now, we have to be cautious and remember that, because <math>2 < e < 3</math> and thus <math>\ln 2 < 1 < \ln 3</math> we have that <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math> and the right-hand side is therefore negative. Since the square on the left-hand side cannot be negative, the equation has no solution. | ||
Current revision
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,
 3ex2 =ln3+lnex2=ln3+x2lne=ln3+x2,=ln2x=xln2. |
If we collect the terms onto one side, the equation becomes
 ln2+ln3=0 |
which is a standard second-order equation for which we complete the square
 x−21ln2 2−21ln22+ln3=0,x−21ln22=21ln22−ln3. |
Now, we have to be cautious and remember that, because 
e31ln3ln3
