From Förberedande kurs i matematik 1

Revision as of 10:28, 2 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_1b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_4_1b-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:54, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,
-	<center> [[Bild:3_4_1b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:3_4_1b-2(2).gif]] </center>	+	Using the log laws, we can divide up the products into several logarithmic terms,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}
		+
		+	and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents
		+
		+	{{Displayed math||<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}
		+
		+	Collect <math>x</math> on one side and the other terms on the other,
		+
		+	{{Displayed math||<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}
		+
		+	Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,
		+
		+	{{Displayed math||<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}
		+
		+	Then, solve for <math>x</math>,
		+
		+	{{Displayed math||<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}
		+
		+
		+	Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as
		+
		+	{{Displayed math||<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}
		+
		+	to indicate that <math>x</math> is negative.

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Current revision

In the equation, both sides are positive because the factors \displaystyle e^{x} and \displaystyle 3^{-x} are positive regardless of the value of \displaystyle x (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,

\displaystyle \ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}

Using the log laws, we can divide up the products into several logarithmic terms,

\displaystyle \ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},

and using the law \displaystyle \ln a^{b}=b\cdot \ln a, we can get rid of \displaystyle x from the exponents

\displaystyle \ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}

Collect \displaystyle x on one side and the other terms on the other,

\displaystyle x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}

Take out \displaystyle x on the left-hand side and use \displaystyle \ln e=1,

\displaystyle x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}

Then, solve for \displaystyle x,

\displaystyle x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}

Note: Because \displaystyle \ln 2 < \ln 13, we can write the answer as

\displaystyle x=-\frac{\ln 13-\ln 2}{1+\ln 3}

to indicate that \displaystyle x is negative.