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Natural Logarithms

Natural logarithms

Natural logarithms are logarithms to the irrational base $e$e where $e\approx2.718281828459$e2.718281828459.

The number $e$e has become one of the most important numbers in the natural sciences, and hence gets its name for that reason. 

In the context of a base, the function given by $y=\log_ex$y=logex has the unique property that a tangent drawn to it at $x=a$x=a will have a gradient of $\frac{1}{a}$1a.

In most modern textbooks, logarithms to base $e$e have been given the special notation of $\ln x$lnx to indicate that it is the logarithm natural of $x$x.


Applying the definition

Recall from the definition of logarithms that if $y=\ln x=\log_ex$y=lnx=logex then $x=e^y$x=ey

So if we substitute $x=e^y$x=ey into $y=\ln x$y=lnx, we see that $y=\ln e^y$y=lney. In other words, raising $e$e to the power $y$y, and then taking logs on that answer restores the original $y$y. In effect they are reverse operations.

If we substitute $y=\ln x$y=lnx into $x=e^y$x=ey, we can also show that $x=e^{\ln x}$x=elnx. Again taking the log of $x$x, and then use that answer as the exponent that $e$e is raised to, simply restores the original $x$x.

Example 1

Using the above argument, the expression $e^{\ln5.4}$eln5.4 is simply $5.4$5.4.

Also the expression $\ln e^{\sqrt{5}}$lne5 is simply $\sqrt{5}$5.


To solve $x\ln\left(\frac{1}{e^5}\right)=x^2+6$xln(1e5)=x2+6 we proceed as follows:

$x\ln\left(\frac{1}{e^5}\right)$xln(1e5) $=$= $x^2+6$x2+6
$x\ln\left(e^{-5}\right)$xln(e5) $=$= $x^2+6$x2+6
$-5x$5x $=$= $x^2+6$x2+6
$x^2+5x+6$x2+5x+6 $=$= $0$0
$\left(x+3\right)\left(x+2\right)$(x+3)(x+2) $=$= $0$0
$x$x $=$= $-3,-2$3,2
Example 3

To solve $\ln\left(\ln e^x\right)=\ln5$ln(lnex)=ln5, we should immediately observe the simplification on the left hand side, so that we are actually solving $\ln\left(x\right)=\ln5$ln(x)=ln5, from which it is clear that $x=5$x=5.

Example 4

To solve $e^{x+\ln9}=2e^{2x}+4$ex+ln9=2e2x+4 we proceed as follows:

$e^{x+\ln9}$ex+ln9 $=$= $2e^{2x}+4$2e2x+4
$e^x\times e^{\ln9}$ex×eln9 $=$= $2e^{2x}+4$2e2x+4
$9e^x$9ex $=$= $2\left(e^x\right)^2+4$2(ex)2+4
$2\left(e^x\right)^2-9\left(e^x\right)+4$2(ex)29(ex)+4 $=$= $0$0

Now set $u=e^x$u=ex and solve the quadratic equation in $u$u

$2u^2-9u+4$2u29u+4 $=$= $0$0
$\left(2u-1\right)\left(u-4\right)$(2u1)(u4) $=$= $0$0
$u$u $=$= $\frac{1}{2},4$12,4
$\therefore$ $e^x$ex $=$= $\frac{1}{2},4$12,4
$\therefore$  $e^x$ex $=$= $\ln\frac{1}{2},\ln4$ln12,ln4
  $=$= $-\ln2,\ln4$ln2,ln4

More Worked Examples

Example 5

Evaluate $\ln\left(\frac{1}{e^2}\right)$ln(1e2).

Example 6

Solve $\ln\left(\ln e^{-y}\right)=\ln3$ln(lney)=ln3.

Example 7

Solve $e^{x+\ln8}=5e^x+3$ex+ln8=5ex+3.



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