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Guided Example #6 - Logarithms to the Base e

In the previous section we studied the natural exponential function with base e, $f (x) = e^{x}$ . In this section we will study its inverse the natural logarithm of x, $f (x) = {log}_{e} x$ , which is also written as ln x. ${log}_{e} x = ln x$ . All the rules for common logarithms apply to the natural logs.

Example

Solve $LaTeX: 4\log_3\left(2e\right)-3\log_e\left(3e\right)$ $4\log_3\left(2e\right)-3\log_e\left(3e\right)$

$LaTeX: =\log_e\left(2e\right)^4-\log_e\left(3e\right)^3$ $=\log_e\left(2e\right)^4-\log_e\left(3e\right)^3$

$LaTeX: =\log_e\left(\frac{\left(2e\right)^4}{\left(3e\right)^3}\right)$ $=\log_e\left(\frac{\left(2e\right)^4}{\left(3e\right)^3}\right)$

$LaTeX: =\log_e\left(\frac{\left(16e^4\right)}{\left(27e^3\right)}\right)$ $=\log_e\left(\frac{\left(16e^4\right)}{\left(27e^3\right)}\right)$

$LaTeX: =\log_e\left(\frac{16e}{27}\right)$ $=\log_e\left(\frac{16e}{27}\right)$

$LaTeX: =\log_ee+\log_e16-\log_e27$ $=\log_ee+\log_e16-\log_e27$

$LaTeX: =1+\log_316-\log_327$ $=1+\log_316-\log_327$

Practice 1:

Solve

\log_{a} (2 p + 1) + \log (3 p - 10) = \log_{a} (11 p)

$\log_a\left(2p+1\right)+\log\left(3p-10\right)=\log_a\left(11p\right)$ for

p > 4

$p>4$

\log_{a} [(2 p + 1) (3 p - 10)] = \log_{a} (11 p)

$\log_a\left[\left(2p+1\right)\left(3p-10\right)\right]=\log_a\left(11p\right)$

(2 p + 1) (3 p - 10) = 11 p

$\left(2p+1\right)\left(3p-10\right)=11p$

6 p^{2} - 28 p - 10 = 0

$6p^2-28p-10=0$

(3 p + 1) (p - 5) = 0

$\left(3p+1\right)\left(p-5\right)=0$

3 p + 1 = 0

$3p+1=0$ or

p - 5 = 0

$p-5=0$

p = - (\frac{1}{3})

$p=-\left(\frac{1}{3}\right)$ or

p = 5

$p=5$

Since

p > 4

$p>4$ , then

p = 5

$p=5$ is the solution.

Practice 2:

Find the inverse of

\log_{3} 9 x

$\log_39x$

Let

y = f_{1} x

$y=f_1x$ , then

x = \log_{3} 9 y

$x=\log_39y$ (interchange x and y)

3^{x} = 9 y

$3^x=9y$

(\frac{3^{x}}{3^{2}}) = y

$\left(\frac{3^x}{3^2}\right)=y$

3^{x - 2} = y

$3^{x-2}=y$

3^{x - 2} = f^{1} (x)

$3^{x-2}=f^1\left(x\right)$