Solution 3.4:2a
From Förberedande kurs i matematik 1
(Difference between revisions)
m (Robot: Automated text replacement (-[[Bild: +[[Image:)) |
m |
||
(2 intermediate revisions not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | The left-hand side is "2 raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\ln 2^{x^2-2} = \ln 1\,,</math>}} |
+ | |||
+ | and use the log law <math>\ln a^b = b\cdot \ln a</math> to get the exponent <math>x^2-2</math> as a factor on the left-hand side, | ||
+ | |||
+ | {{Displayed math||<math>\bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}</math>}} | ||
+ | |||
+ | Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving | ||
+ | |||
+ | {{Displayed math||<math>(x^2-2)\ln 2=0\,\textrm{.}</math>}} | ||
+ | |||
+ | This means that ''x'' must satisfy the second-degree equation | ||
+ | |||
+ | {{Displayed math||<math>x^2-2 = 0\,\textrm{.}</math>}} | ||
+ | |||
+ | Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>. | ||
+ | |||
+ | |||
+ | Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007. |
Current revision
The left-hand side is "2 raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,
\displaystyle \ln 2^{x^2-2} = \ln 1\,, |
and use the log law \displaystyle \ln a^b = b\cdot \ln a to get the exponent \displaystyle x^2-2 as a factor on the left-hand side,
\displaystyle \bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.} |
Because \displaystyle e^{0}=1, so \displaystyle \ln 1 = 0, giving
\displaystyle (x^2-2)\ln 2=0\,\textrm{.} |
This means that x must satisfy the second-degree equation
\displaystyle x^2-2 = 0\,\textrm{.} |
Taking the root gives \displaystyle x=-\sqrt{2} or \displaystyle x=\sqrt{2}\,.
Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.