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Solution 3.4:1b

From Förberedande kurs i matematik 1

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Current revision (08:54, 2 October 2008) (edit) (undo)
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In the equation, both sides are positive because the factors
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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,
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<math>e^{x}</math>
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and
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<math>3^{-x}</math>
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are positive regardless of the value of
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<math>x</math>
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(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,
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{{Displayed math||<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}
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<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>
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Using the log laws, we can divide up the products into several logarithmic terms,
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{{Displayed math||<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}
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Using the log law, we can divide up the products into several logarithmic terms,
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and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents
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{{Displayed math||<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}
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<math>\ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}</math>
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Collect <math>x</math> on one side and the other terms on the other,
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{{Displayed math||<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}
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and using the law
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Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,
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<math>\ln a^{b}=b\centerdot \ln a</math>, we can get rid of
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<math>x</math>
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from the exponents:
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{{Displayed math||<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}
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<math>\ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3</math>
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Then, solve for <math>x</math>,
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{{Displayed math||<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}
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Collecting together
 
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<math>x</math>
 
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on one side and the other terms on the other,
 
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Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as
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<math>x\ln e+x\ln 3=\ln 2-\ln 13</math>
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{{Displayed math||<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}
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to indicate that <math>x</math> is negative.
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Take out
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<math>x</math>
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on the left-hand side and use
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<math>\ln e=1</math>
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:
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<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>
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Then, solve for
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<math>x</math>
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:
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<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>
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NOTE: Because
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<math>\ln 2<\ln 13</math>, we can write the answer as
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<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>
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in order to indicate that
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<math>x</math>
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is negative.
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Current revision

In the equation, both sides are positive because the factors ex and 3−x are positive regardless of the value of x (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,

ln13ex=ln23−x. 

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

ln13+lnex=ln2+ln3−x

and using the law lnab=blna, we can get rid of x from the exponents

ln13+xlne=ln2+(−x)ln3.

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

xlne+xln3=ln2−ln13.

Take out x on the left-hand side and use lne=1,

x(1+ln3)=ln2−ln13.

Then, solve for x,

x=1+ln3ln2−ln13.


Note: Because ln2ln13, we can write the answer as

x=−1+ln3ln13−ln2

to indicate that x is negative.

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