03Flogeqnsbase(e)

toc = Equations with natural (base e) logarithms =

The logarithm laws apply to log e (x) {or ln(x)} as with any other base.

math e^x=a \; \; \iff \;\; \log_ea=x \; \textbf{ or } \; ln(a)=x math

**Example 1**
Solve for //x//, answer both in exact form and correct to 3 decimal places:

math \\ (a) \quad \log_e{25}=x \\ \\ (b) \quad \log_ex=2 math

__Solution:__

Solution is already in exact form, but it can be simplified:
 * (a) **

math \\ x = \log_e{25} \\ \\ x= \log_e5^2 \\ \\ x=2 \log_e5 \qquad \textit{exact value} math math x=3.219 \qquad \textit{corrrect to 3 decimal places} math

math \log_ex=2 \iff x=e^2 \qquad \textit{exact value} math
 * (b) **

math x=7.389 \qquad \textit{correct to 3 decimal places} math

**Example 2**
Solve for x: math \\ (a) \quad \log_e3x= \log_e72 \\ \\ (b) \quad \log_ex+ \log_e(5x-12)= \log_e81 math

__Solution:__

math \log_e3x=\log_e72 math We have log(a) = log(b), hence a = b math \\ 3x=72 \\ \\ x=24 math math \log_ex+ \log_e(5x-12)= \log_e81 math Use log laws to combine left side into one log before we can equate the insides math \log_e{x(5x-12)}= \log_e81 math
 * (a) **
 * (b) **

math \\ x(5x-12)=81 \\ \\ 5x^2-12x-81=0 \\ \\ (5x-27)(x+3)=0 \\ \\ 5x-27=0 \;\; \textit{ or } \;\; x+3=0 \\ \\ x=\frac{27}{5} \qquad \quad \text{or } \;\; x = -3 math

Reject x = –3 as a solution because x > 0 for log a (x) to be defined

hence math x=\dfrac{27}{5} math Go to top of page flat .