Solution 3.4:1c
From Förberedande kurs i matematik 1
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- | The equation has the same form as the equation in exercise | + | 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, | First, we take logs of both sides, | ||
+ | {{Displayed math||<math>\ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}</math>}} | ||
- | <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, | |
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- | Then, collect together the <math>x</math> terms on the left-hand side | + | |
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+ | {{Displayed math||<math>x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}</math>}} | ||
The solution is now | 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>}} | |
- | <math>x=\frac{\ln 7-\ln 3}{\ln e-\ln 2}=\frac{\ln 7-\ln 3}{1-\ln 2}</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{.} |