Operations for functions: Exponential and logarithmic functions
Properties of the exponential functions
Properties of the exponents
Let #a# and #b# be two positive real numbers. The following rules of calculation hold for the exponential functions #a^x# and #b^x#.
| 1 | \(a^0=1\) |
| 2 | \(a^{x+y}=a^x\cdot a^y\) |
| 3 | \(a^{-x}=\frac{1}{a^x}=\left (\frac{1}{a} \right )^x\) |
| 4 | \(a^{x-y}=\frac{a^x}{a^y}\) |
| 5 | \(\left(a^x\right)^y = a^{x\cdot y}\) |
| 6 | \((a\cdot b)^x = a^x\cdot b^x\) |
| 7 | if #a\lt b# and #x\gt 0#, then #a^x\lt b^x# |
| 8 | if #a\gt 1# and #x\lt y#, then #a^x\lt a^y# |
In particular, #a^x# is increasing if #a\gt1# and decreasing if #a\lt 1#.
These rules are direct consequences of the known calculation rules for arithmetic.
By way of example, we use the rules to show why #a^x# is decreasing if #a\lt 1#. Suppose #x# and #y# are real numbers with #x\lt y#. Then we have:
\[ \begin{array}{rcl}\left(\frac{1}{a}\right)^x&\lt& \left(\frac{1}{a}\right)^y\\ &&\phantom{xx}\color{blue}{\text{rule 8}}\\ \left(\frac{1}{a}\right)^x\cdot a^{x+y}&\lt& \left(\frac{1}{a}\right)^y\cdot a^{x+y}\\ &&\phantom{xx}\color{blue}{\text{both sides multiplied by the same positive number}}\\ {a}^{-x}\cdot a^{x+y}&\lt& {a}^{-y}\cdot a^{x+y}\\ &&\phantom{xx}\color{blue}{\text{rule 3}}\\ a^{y}&\lt& a^{x}\\ &&\phantom{xx}\color{blue}{\text{rule 2}} \end{array}\]This shows that #a^x\gt a^y#, so #a^x# is decreasing.
Here are some examples:
| 1 | \(2^0=1\) |
| 2 | \(3^{x+y}=3^3\cdot 3^y\) |
| 3 | \(5^{-x}=\frac{1}{5^x}=\left (\frac{1}{5} \right )^x\) |
| 4 | \(4^{2}=\frac{4^5}{4^3}\) |
| 5 | \(\left(2^5\right)^6 = 2^{30}\) |
| 6 | \( 6^x = 2^x\cdot 3^x \) |
From #a^x\cdot a^y=a^{x+y}# we deduce #a^4\cdot a^8=a^{4+8}=a^{12}#.
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