Math Problem Statement

1. Bukti eksponen, 2. Tentukan nilai p sedemikian hingga persamaan berikut tepat, 3. Sederhanakanlah beberapa ekspresi matematika.

Solution

Let's go through the questions in this image one by one:


1. Prove the exponent properties for numbers 6 and 7.

You need to refer to the exponent properties, such as:

  • For 6, likely am×an=am+na^m \times a^n = a^{m+n}
  • For 7, likely (am)n=am×n(a^m)^n = a^{m \times n}

For the proof, you should use the definition of exponents and basic algebraic manipulation.


2. Determine the value of pp so that the following equations hold true:

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

  • Using the exponent rule (am)n=am×n(a^m)^n = a^{m \times n}, the left-hand side becomes: (34)2=34×2=38(3^4)^2 = 3^{4 \times 2} = 3^8 Therefore, p=8p = 8.

b. bp×b5=b9b^p \times b^5 = b^9

  • Using the rule am×an=am+na^m \times a^n = a^{m+n}, the left-hand side becomes: bp×b5=bp+5b^p \times b^5 = b^{p+5} For bp+5=b9b^{p+5} = b^9, we solve: p+5=9p=4p + 5 = 9 \quad \Rightarrow \quad p = 4

c. (3π)p=27π3(3\pi)^p = 27\pi^3

  • Rewrite 27π327\pi^3 as (33)π3(3^3)\pi^3, so the equation becomes: (3π)p=33×π3(3\pi)^p = 3^3 \times \pi^3 This suggests that p=3p = 3 because the exponents of 33 and π\pi match.

3. Simplify the following expressions:

a. (24×3623×32)3\left( \frac{2^4 \times 3^6}{2^3 \times 3^2} \right)^3

  • First, simplify inside the parentheses: 24×3623×32=243×362=21×34\frac{2^4 \times 3^6}{2^3 \times 3^2} = 2^{4-3} \times 3^{6-2} = 2^1 \times 3^4 Now, apply the exponent rule (am)n=am×n(a^m)^n = a^{m \times n}: (21×34)3=21×3×34×3=23×312(2^1 \times 3^4)^3 = 2^{1 \times 3} \times 3^{4 \times 3} = 2^3 \times 3^{12}

b. (3u3v5)(9u4v)(3u^3v^5)(9u^4v)

  • Simplify by multiplying the constants and adding the exponents for the same bases: 3×9=273 \times 9 = 27 For uu, u3×u4=u3+4=u7u^3 \times u^4 = u^{3+4} = u^7, and for vv, v5×v1=v5+1=v6v^5 \times v^1 = v^{5+1} = v^6. Thus, the result is: 27u7v627u^7v^6

c. (n1r45n6r4)2\left( \frac{n^{-1} r^4}{5n^{-6} r^4} \right)^2 where n0,r0n \neq 0, r \neq 0

  • Simplify inside the parentheses: n1r45n6r4=n15n6=n1n6×5=n1×n65=n55\frac{n^{-1} r^4}{5n^{-6} r^4} = \frac{n^{-1}}{5n^{-6}} = \frac{n^{-1}}{n^{-6} \times 5} = \frac{n^{-1} \times n^6}{5} = \frac{n^{5}}{5} Now, apply the exponent outside: (n55)2=n5×252=n1025\left( \frac{n^{5}}{5} \right)^2 = \frac{n^{5 \times 2}}{5^2} = \frac{n^{10}}{25}

Would you like further details or have any questions on these problems? Below are five related questions to expand on this:

  1. Can you prove the exponent rule am×an=am+na^m \times a^n = a^{m+n} with a general example?
  2. How do you solve equations involving logarithms if exponents are not simple integers?
  3. How can you simplify more complex expressions involving multiple bases and exponents?
  4. What are some applications of exponent rules in real-world problems?
  5. How do you handle equations with negative or fractional exponents?

Tip: Always check the base in exponent rules to make sure they are the same before applying properties.

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Math Problem Analysis

Mathematical Concepts

Exponentiation
Algebraic simplification

Formulas

(a^m)^n = a^{m \times n}
a^m \times a^n = a^{m+n}
(a^m)/(a^n) = a^{m-n}

Theorems

Exponentiation rules
Power of a power rule

Suitable Grade Level

Grades 10-12