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Exponents and Logarithms

Property Proofs

Nabil Akbarazzima Fatih

Mathematics

Exponential Properties and Their Proofs

We will discuss the properties of exponents and their proofs. These properties are essential to understand as they form the foundation for solving various problems related to exponents and logarithms.

Property 1

am⋅an=am+na^m \cdot a^n = a^{m+n}

For every a≠0a \neq 0 and m,nm, n real numbers.

Property 2

aman=am−n\frac{a^m}{a^n} = a^{m-n}

For every a≠0a \neq 0 and m,nm, n real numbers.

Property 3

(am)n=am⋅n(a^m)^n = a^{m \cdot n}

For every a≠0a \neq 0 and m,nm, n real numbers.

Property 4

(ab)m=am⋅bm(ab)^m = a^m \cdot b^m

For every a,b≠0a, b \neq 0 and mm integer.

Proof:

Method 1:

(ab)m=ab×ab×ab×…×ab⏟m factors(ab)^m = \underbrace{ab \times ab \times ab \times \ldots \times ab}_{m\text{ factors}}
(ab)m=a×a×a×…×a⏟m factors×b×b×b×…×b⏟m factors(ab)^m = \underbrace{a \times a \times a \times \ldots \times a}_{m\text{ factors}} \times \underbrace{b \times b \times b \times \ldots \times b}_{m\text{ factors}}
(ab)m=am×bm(ab)^m = a^m \times b^m

Method 2:

We can take several examples with specific values of m,am, a and bb to observe the pattern that forms. For example, with m=2,a=3,b=4m=2, a=3, b=4:

(3⋅4)2=122=144(3 \cdot 4)^2 = 12^2 = 144
32⋅42=9⋅16=1443^2 \cdot 4^2 = 9 \cdot 16 = 144

Property 5

(ab)m=ambm\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}

For every a≠0,b≠0a \neq 0, b \neq 0 and mm integer.

Proof:

(ab)m=ab×ab×ab×…×ab⏟m factors\left(\frac{a}{b}\right)^m = \underbrace{\frac{a}{b} \times \frac{a}{b} \times \frac{a}{b} \times \ldots \times \frac{a}{b}}_{m\text{ factors}}
(ab)m=a×a×a×…×a⏟m factorsb×b×b×…×b⏟m factors\left(\frac{a}{b}\right)^m = \frac{\underbrace{a \times a \times a \times \ldots \times a}_{m\text{ factors}}}{\underbrace{b \times b \times b \times \ldots \times b}_{m\text{ factors}}}
(ab)m=ambm\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}

Property 6

(amn)(apn)=am+pn(a^{\frac{m}{n}})(a^{\frac{p}{n}}) = a^{\frac{m+p}{n}}

For every a>0a > 0, mn\frac{m}{n} and pn\frac{p}{n} rational numbers with n≠0n \neq 0.

Proof:

Using Property 1, we obtain:

(amn)(apn)=amn+pn(a^{\frac{m}{n}})(a^{\frac{p}{n}}) = a^{\frac{m}{n} + \frac{p}{n}}
(amn)(apn)=am+pn(a^{\frac{m}{n}})(a^{\frac{p}{n}}) = a^{\frac{m+p}{n}}

Property 7

(amn)(apq)=amq+pnnq(a^{\frac{m}{n}})(a^{\frac{p}{q}}) = a^{\frac{mq + pn}{nq}}

For every a>0a > 0, mn\frac{m}{n} and pq\frac{p}{q} rational numbers with n,q≠0n, q \neq 0.

Proof:

Using Property 1, we need to equalize the denominators of the exponents:

(amn)(apq)=amn+pq(a^{\frac{m}{n}})(a^{\frac{p}{q}}) = a^{\frac{m}{n} + \frac{p}{q}}
(amn)(apq)=amqnq+pnnq(a^{\frac{m}{n}})(a^{\frac{p}{q}}) = a^{\frac{mq}{nq} + \frac{pn}{nq}}
(amn)(apq)=amq+pnnq(a^{\frac{m}{n}})(a^{\frac{p}{q}}) = a^{\frac{mq + pn}{nq}}

Example Problems

Determining Values and Proofs

  1. (34)2=3p(3^4)^2 = 3^p

    Solution: Using Property 3, (am)n=am⋅n(a^m)^n = a^{m \cdot n}

    (34)2=34⋅2=38(3^4)^2 = 3^{4 \cdot 2} = 3^8
    p=8p = 8
  2. b4⋅b5=b9b^4 \cdot b^5 = b^9

    Solution: Using Property 1, am⋅an=am+na^m \cdot a^n = a^{m+n}

    b4⋅b5=b4+5=b9b^4 \cdot b^5 = b^{4+5} = b^9
    Therefore, the equation is proven true.\text{Therefore, the equation is proven true.}
  3. (3π)p=27π3(3\pi)^p = 27\pi^3

    Solution: Converting 27π327\pi^3 to 33⋅π33^3 \cdot \pi^3

    27π3=33⋅π3=(3π)327\pi^3 = 3^3 \cdot \pi^3 = (3\pi)^3
    p=3p = 3

Simplifying Expressions

  1. (24×3623×32)3\left(\frac{2^4 \times 3^6}{2^3 \times 3^2}\right)^3
    (24×3623×32)3=(24−3×36−2)3\left(\frac{2^4 \times 3^6}{2^3 \times 3^2}\right)^3 = \left(2^{4-3} \times 3^{6-2}\right)^3
    =(21×34)3= \left(2^1 \times 3^4\right)^3
    =21⋅3×34⋅3= 2^{1 \cdot 3} \times 3^{4 \cdot 3}
    =23×312= 2^3 \times 3^{12}
  2. (3u3v5)(9u4v)(3u^3v^5)(9u^4v)
    (3u3v5)(9u4v)=3⋅9⋅u3+4⋅v5+1(3u^3v^5)(9u^4v) = 3 \cdot 9 \cdot u^{3+4} \cdot v^{5+1}
    =31⋅32⋅u7⋅v6= 3^1 \cdot 3^2 \cdot u^7 \cdot v^6
    =31+2⋅u7⋅v6= 3^{1+2} \cdot u^7 \cdot v^6
    =33⋅u7⋅v6= 3^3 \cdot u^7 \cdot v^6
    =27u7v6= 27u^7v^6
  3. (n−1r45n−6r−4)2\left(\frac{n^{-1}r^4}{5n^{-6}r^{-4}}\right)^2
    (n−1r45n−6r−4)2=(n−1r45⋅1n−6r−4)2\left(\frac{n^{-1}r^4}{5n^{-6}r^{-4}}\right)^2 = \left(\frac{n^{-1}r^4}{5} \cdot \frac{1}{n^{-6}r^{-4}}\right)^2
    =(n−1r45⋅n6r4)2= \left(\frac{n^{-1}r^4}{5} \cdot n^6r^4\right)^2
    =(n−1+6r4+45)2= \left(\frac{n^{-1+6}r^{4+4}}{5}\right)^2
    =(n5r85)2= \left(\frac{n^5r^8}{5}\right)^2
    =n10r1625= \frac{n^{10}r^{16}}{25}
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