Inverse of exponential functions
In mathematics, inverse functions can be thought of as opposites of each other, like opposite operations, e.g. addition & subtraction, multiplication & division. The inverse function of an exponential is called a logarithm, and it works like this:
If we have an exponential of the form:
$b^x=a$bx=a
this can be re-expressed as the logarithm:
$\log_ba=x$logba=x
and $b\ne1$b≠1 is called the base.
In other words, logarithms are useful when we are interested in finding the exponent needed ($x$x) to raise a certain base ($b$b) to a certain number ($a$a).
Base $10$10 logarithms, are also known as common logarithms and were originally looked up in a table of logs, prior to being able to evaluate them using calculators. Since they are such a commonly used base, the notation for them was simplified and $\log_{10}a$log10a became simply $\log a$loga.
Recognising that logarithms and exponents are inverses can also be very useful when simplifying problems. From the definition of logarithms, if $y=\log_ax$y=logax then $x=a^y$x=ay for any base $a$a.
Substituting $x=a^y$x=ay into $y=\log_ax$y=logax, notice that $y=\log_aa^y$y=logaay. Raising $a$a to the power $y$y, and then taking logs on that answer restores the original $y$y. In effect, they are reverse operations (they "undo" each other).
The reverse is true as well. If $y=\log_ax$y=logax is substituted into $x=a^y$x=ay, this gives $x=a^{\log_ax}$x=alogax. Again, taking the $\log_a$loga of $x$x, and then using that answer as the exponent that $a$a is raised to restores the original $x$x.
The following are very useful for simplifying expressions involving both logarithms and exponents for any base $a$a:
$a^{\log_ax}=x$alogax=x
$\log_aa^x=x$logaax=x
Worked examples
example 1
Express $6^2=36$62=36 in logarithmic form
Think: Remember we want the exponent on its own and the logarithm on the other side.
Do:
$6$6 is the base and $2$2 is the exponent, so:
$\log_636=2$log636=2
example 2
Rewrite in exponential form: $\log_432=2.5$log432=2.5
Think: Remember that in exponential form, we want to isolate the resulting number after having the base raised to the exponent
Do:
$32$32 is the result after raising the base $4$4 to the exponent $2.5$2.5, so:
$4^{2.5}=32$42.5=32
Practice question
question 1
Evaluate $\log_216$log216.
Question 2
Rewrite the equation $\log_3\left(\frac{1}{3}\right)=-1$log3(13)=−1 in exponential form.
Question 3
A function is defined as $f\left(x\right)=\log_2\left(x-2\right)$f(x)=log2(x−2).
Evaluate $f\left(10\right)$f(10).
Natural logarithms
aNatural logarithms are logarithms to the irrational base $e$e (it cannot be written as a simple fraction) where $e\approx2.718281828459$e≈2.718281828459. The number $e$e has become one of the most important numbers in the natural sciences and is often called Euler's number after mathematician Leonhard Euler (pronounced "Oiler").
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, substituting $x=e^y$x=ey into $y=\ln x$y=lnx, notice 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 (they "undo" each other).
If $y=\ln x$y=lnx is substituted into $x=e^y$x=ey, this gives $x=e^{\ln x}$x=elnx. Again, taking the log of $x$x, and then using that answer as the exponent that $e$e is raised to, simply restores the original $x$x. Note that this is a specific application of the general case considered in the previous section where the base is $e$e.
The following are very useful for simplifying expressions involving the natural logarithm and $e$e:
$e^{\ln a}=a$elna=a
$\ln e^b=b$lneb=b
Worked examples
Example 3
Using the above argument, the expression $e^{\ln5.4}$eln5.4 is simply $5.4$5.4.
Also the expression $\ln e^{\sqrt{5}}$lne√5 is simply $\sqrt{5}$√5.
EXAMPLE 4
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(e−5) | $=$= | $x^2+6$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 5
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 6
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)2−9(ex)+4 | $=$= | $0$0 |
Now set $u=e^x$u=ex and solve the quadratic equation in $u$u.
| $2u^2-9u+4$2u2−9u+4 | $=$= | $0$0 |
| $\left(2u-1\right)\left(u-4\right)$(2u−1)(u−4) | $=$= | $0$0 |
| $u$u | $=$= | $\frac{1}{2},4$12,4 |
| $\therefore$∴ $e^x$ex | $=$= | $\frac{1}{2},4$12,4 |
| $\therefore$∴ $x$x | $=$= | $\ln\frac{1}{2},\ln4$ln12,ln4 |
| $=$= | $-\ln2,\ln4$−ln2,ln4 |
Practice question
Question 4
Evaluate $\ln\left(\frac{1}{e^2}\right)$ln(1e2).
Question 5
Solve $\ln\left(\ln e^{-y}\right)=\ln3$ln(lne−y)=ln3.
question 6
Solve $e^{x+\ln8}=5e^x+3$ex+ln8=5ex+3.