From Förberedande kurs i matematik 1

Revision as of 10:29, 2 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_1c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (10:50, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The equation has the same form as the equation in exercise b and we can therefore use the same strategy.
-	<center> [[Bild:3_4_1c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	First, we take logs of both sides,
		+
		+	{{Displayed math||<math>\ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}</math>}}
		+
		+	and use the log laws to make <math>x</math> more accessible,
		+
		+	{{Displayed math||<math>\ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}</math>}}
		+
		+	Then, collect together the <math>x</math> terms on the left-hand side,
		+
		+	{{Displayed math||<math>x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}</math>}}
		+
		+	The solution is now
		+
		+	{{Displayed math||<math>x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}</math>}}

---

Current revision

The equation has the same form as the equation in exercise b and we can therefore use the same strategy.

First, we take logs of both sides,

\displaystyle \ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}

and use the log laws to make \displaystyle x more accessible,

\displaystyle \ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}

Then, collect together the \displaystyle x terms on the left-hand side,

\displaystyle x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}

The solution is now

\displaystyle x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}