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Exponents and Logarithms

Property Proofs

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}am⋅an=am+n

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

Property 2

aman=am−n\frac{a^m}{a^n} = a^{m-n}anam​=am−n

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

Property 3

(am)n=am⋅n(a^m)^n = a^{m \cdot n}(am)n=am⋅n

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

Property 4

(ab)m=am⋅bm(ab)^m = a^m \cdot b^m(ab)m=am⋅bm

For every a,b≠0a, b \neq 0a,b=0 and mmm 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=m factorsab×ab×ab×…×ab​​
(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=m factorsa×a×a×…×a​​×m factorsb×b×b×…×b​​
(ab)m=am×bm(ab)^m = a^m \times b^m(ab)m=am×bm

Method 2:

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

(3⋅4)2=122=144(3 \cdot 4)^2 = 12^2 = 144(3⋅4)2=122=144
32⋅42=9⋅16=1443^2 \cdot 4^2 = 9 \cdot 16 = 14432⋅42=9⋅16=144

Property 5

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

For every a≠0,b≠0a \neq 0, b \neq 0a=0,b=0 and mmm 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}}(ba​)m=m factorsba​×ba​×ba​×…×ba​​​
(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}}}(ba​)m=m factorsb×b×b×…×b​​m factorsa×a×a×…×a​​​
(ab)m=ambm\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}(ba​)m=bmam​

Property 6

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

For every a>0a > 0a>0, mn\frac{m}{n}nm​ and pn\frac{p}{n}np​ rational numbers with n≠0n \neq 0n=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}}(anm​)(anp​)=anm​+np​
(amn)(apn)=am+pn(a^{\frac{m}{n}})(a^{\frac{p}{n}}) = a^{\frac{m+p}{n}}(anm​)(anp​)=anm+p​

Property 7

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

For every a>0a > 0a>0, mn\frac{m}{n}nm​ and pq\frac{p}{q}qp​ rational numbers with n,q≠0n, q \neq 0n,q=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}}(anm​)(aqp​)=anm​+qp​
(amn)(apq)=amqnq+pnnq(a^{\frac{m}{n}})(a^{\frac{p}{q}}) = a^{\frac{mq}{nq} + \frac{pn}{nq}}(anm​)(aqp​)=anqmq​+nqpn​
(amn)(apq)=amq+pnnq(a^{\frac{m}{n}})(a^{\frac{p}{q}}) = a^{\frac{mq + pn}{nq}}(anm​)(aqp​)=anqmq+pn​

Example Problems

Determining Values and Proofs

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

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

    (34)2=34⋅2=38(3^4)^2 = 3^{4 \cdot 2} = 3^8(34)2=34⋅2=38
    p=8p = 8p=8
  2. b4⋅b5=b9b^4 \cdot b^5 = b^9b4⋅b5=b9

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

    b4⋅b5=b4+5=b9b^4 \cdot b^5 = b^{4+5} = b^9b4⋅b5=b4+5=b9
    Therefore, the equation is proven true.\text{Therefore, the equation is proven true.}Therefore, the equation is proven true.
  3. (3π)p=27π3(3\pi)^p = 27\pi^3(3π)p=27π3

    Solution: Converting 27π327\pi^327π3 to 33⋅π33^3 \cdot \pi^333⋅π3

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

Simplifying Expressions

  1. (24×3623×32)3\left(\frac{2^4 \times 3^6}{2^3 \times 3^2}\right)^3(23×3224×36​)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(23×3224×36​)3=(24−3×36−2)3
    =(21×34)3= \left(2^1 \times 3^4\right)^3=(21×34)3
    =21⋅3×34⋅3= 2^{1 \cdot 3} \times 3^{4 \cdot 3}=21⋅3×34⋅3
    =23×312= 2^3 \times 3^{12}=23×312
  2. (3u3v5)(9u4v)(3u^3v^5)(9u^4v)(3u3v5)(9u4v)
    (3u3v5)(9u4v)=3⋅9⋅u3+4⋅v5+1(3u^3v^5)(9u^4v) = 3 \cdot 9 \cdot u^{3+4} \cdot v^{5+1}(3u3v5)(9u4v)=3⋅9⋅u3+4⋅v5+1
    =31⋅32⋅u7⋅v6= 3^1 \cdot 3^2 \cdot u^7 \cdot v^6=31⋅32⋅u7⋅v6
    =31+2⋅u7⋅v6= 3^{1+2} \cdot u^7 \cdot v^6=31+2⋅u7⋅v6
    =33⋅u7⋅v6= 3^3 \cdot u^7 \cdot v^6=33⋅u7⋅v6
    =27u7v6= 27u^7v^6=27u7v6
  3. (n−1r45n−6r−4)2\left(\frac{n^{-1}r^4}{5n^{-6}r^{-4}}\right)^2(5n−6r−4n−1r4​)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(5n−6r−4n−1r4​)2=(5n−1r4​⋅n−6r−41​)2
    =(n−1r45⋅n6r4)2= \left(\frac{n^{-1}r^4}{5} \cdot n^6r^4\right)^2=(5n−1r4​⋅n6r4)2
    =(n−1+6r4+45)2= \left(\frac{n^{-1+6}r^{4+4}}{5}\right)^2=(5n−1+6r4+4​)2
    =(n5r85)2= \left(\frac{n^5r^8}{5}\right)^2=(5n5r8​)2
    =n10r1625= \frac{n^{10}r^{16}}{25}=25n10r16​
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Exponent Properties

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Function Exploration

  • Property ProofsLearn rigorous proofs of 7 exponential properties with step-by-step demonstrations. Master multiplication, division, and power rules through examples.
On this page
  • Exponential Properties and Their Proofs
    • Property 1
    • Property 2
    • Property 3
    • Property 4
    • Property 5
    • Property 6
    • Property 7
  • Example Problems
    • Determining Values and Proofs
    • Simplifying Expressions
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