04 21logeqnse

Equations with natural (base e) logarithms

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Recall that e is Euler's number. $e \approx 2.71828$
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The logarithm laws apply to $\log_e(x)$ as with any other base.

• $\log_e(x)$ can also be written as $\text{ln}(x)$. Either notation is acceptable.
• You need to be familiar with both forms.

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… … $e^x=a \; \; \iff \;\; \log_e(a)=x$

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Example 1

Solve for x, answer both in exact form and correct to 3 decimal places:
a)$\log_e(25)=x$

b)$\log_e(x)=2$

Solution:

a)$x = \log_e(25)$

… … … $x= \log_e \big(5^2\big)$

… … … $x=2 \log_e(5)$ … … {exact value}

… … … $x=3.219$ … … {corrrect to 3 decimal places}
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b)$\log_e(x)=2 \iff x=e^2$

… … … $x= e^2$ … … {exact value}

… … … $x=7.389$ … … {correct to 3 decimal places}

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Example 2

Solve for x:
a)$\log_e(3x) = \log_e(72)$

b) … \log_e(x) + \log_e(5x-12)= \log_e(81)

Solution:

a)$\log_e(3x) = \log_e(72)$

… … … We have log(a) = log(b), hence a = b

… … … $3x=72$

… … … $x=24$
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b)$\log_e(x) + \log_e(5x-12)= \log_e(81)$

… … … Use log laws to combine left side into one log

… … … $\log_e \big( x(5x-12)\big)= \log_e(81)$

… … … $x(5x-12)=81$

… … … $5x^2-12x-81=0$

… … … $(5x-27)(x+3)=0$

… … … $5x-27=0 \;\; \text{ or } \;\; x+3=0$

… … … $x=\frac{27}{5} \qquad \quad \text{ or } \;\; x = -3$

… … … Reject $x = -3$ as a solution because x > 0 for loga(x) to be defined

… … … hence

… … … $x=\dfrac{27}{5}$

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