lembaranbaru.my.id – The logarithm problem for 10th grade students here can be used to measure students’ understanding. Studying a collection of logarithm problems for 10th grade along with their answers is also useful for preparation before exams.
Logarithm is a mathematical operation that is the inverse or opposite of exponents. This means that to find the value of a logarithmic number, you need to reverse its exponential function. The same goes for logarithmic functions.
Learning with practice logarithm problems for 10th grade will make it easier for students to understand the content of this Mathematics material. Therefore, check out the list of examples of 10th grade logarithm problems from the Independent Curriculum below.
10th Grade Logarithm Problems Along with Their Answers
10th grade logarithm problems can come in the form of equations or stories. Of course, the level of difficulty varies. Here are 30 examples of 10th grade logarithm problems along with their answers:
1. Find the value of ²log 24 – ²log 3!
Answer: The problem above can be solved with the following calculation…
- ²log 24 – ²log 3
- = ²log 24/3
- = ²log 8
- = ²log 2³
- = 3
So, the value of ²log 24 – ²log 3 is 3.
2. Simplify the logarithmic form of ²log 8 + ²log 4!
Answer: The simplification of the form above uses the addition property of logarithms, which is logₐ x + logₐ y = log (x.y). Then, use the logarithmic power property, which is logₐ (xⁿ) = n x logₐ x.
The solution is:
- ²log 8 + ²log 4
- = ²log (8 x 4)
- = ²log 32
- = ²log 2⁵
- = 5 x ²log 2
- = 5 x 1 = 5.
3. Given 5log 3 = a and 3log 4 = b, determine 12log 75 in terms of a and b!
Answer:
- 12log 75 = 3log 75/3log 12 = 3log (3 × 25)/3log (3 × 4)
- 3log (3 × 25)/3log (3 × 4) = (3log 3 + 3log 52)/(3log 3 + 3log 4)
- (3log 3 + 3log 52)/(3log 3 + 3log 4) = (1 + 2 × 3log 5)/(1 +b) = (1 + 2 × 1/a)/(1 +b) = a + 2/(1+b)a
4. Simplify the logarithmic form of alog 1/b × blog 1/c × clog 1/a!
Answer:
- alog 1/b × blog 1/c × clog 1/a
- = alog b-1 × blog c-1 × clog a-1
- = –alog b × -blog c × -clog a
- = -alog b × blog c × clog a
- = –alog a
- = -1
5. Determine the value of ²log 4 + ²log 12 – ²log 6! Explain!
Answer:
We use the logarithmic property “ᵃlog(b.c) = ᵃlog b + ᵃlog c” AND “ᵃlog b/c = ᵃlog b – ᵃlog c.”
The solution is as follows:
²log 4 + ²log 12 – ²log 6 = ²log (4 x 12) / 6 = ²log 8
Then, applying the logarithmic property “ᵃlog bⁿ = n x ᵃlog b” to find the final result. So, the solution to find the result is:
- ²log 8
- = ²log 2³ (because 8 = 2³)
- = 3 x ²log 2
- = 3 x 1 (because ᵃlog a = 1)
- = 3.
Therefore, the logarithmic value of ²log 4 + ²log 12 – ²log 6 is 3.
6. Determine the value of ²log 48 + ⁵log 50 – ²log 3 – ⁵log 2! Explain!
Answer: Since the logarithm indexes are not uniform (there are ²log and ⁵log), the solution needs to involve rearranging the addition and subtraction operations in the problem.
The problem above can be rearranged as ²log 48 – ²log 3 + ⁵log 50 – ⁵log 2.
Therefore, simplification can use the logarithmic property “ᵃlog b/c = ᵃlog b – ᵃlog c.”
The solution is as follows:
- ²log 48 – ²log 3 + ⁵log 50 – ⁵log 2
- = (²log 48/3) + (⁵log 50/2)
- = ²log 16 + ⁵log 25
To find the result, you can use the logarithmic property “ᵃlog a = 1” AND “ᵃlog bⁿ = n x ᵃlog b.”
Therefore, the solution to find the result is:
- ²log 16 + ⁵log 25
- = ²log 2⁴ + ⁵log 5² [because 16 = 2⁴ and 25 = 5²]
- = (4 x ²log 2) + (2 x ⁵log 5)
- = 4 + 2
- = 6
Therefore, the logarithmic value of ²log 48 + ⁵log 50 – ²log 3 – ⁵log 2 is 6.
7. Determine the value of ²log (125/9), if it is known that ²log 3 = 1.6 and ²log 5 = 2.3! Explain!
Answer: The solution can follow several steps as follows.
- a. Since 125 = 5³ and 9 = 3², then ²log (125/9) = ²log (5³/3²)
- b. By using the logarithmic property “ᵃlog b/c = ᵃlog b – ᵃlog c,” ²log (5³/3²) = ²log 5³ – ²log 3²
- c. To find the final result, you can apply the logarithmic property “ᵃlog bⁿ = n x ᵃlog b”
- d. The solution is as follows:
- ²log 5³ – ²log 3²
- = (3 x ²log 5) – (2 x ²log 3)
- = (3 x 2.3) – (2 x 1.6)
- = 6.9 – 3.2
- = 3.7
Therefore, since it is known that ²log 3 = 1.6 and ²log 5 = 2.3, the value of ²log (125/9) is 3.7.
8. In chemistry, the acidity level of a solution is called pH. The acidity level can be defined with the logarithmic function p(t) = -log t, with t being the concentration of hydrogen ions (+H) expressed in moles per liter (mol/L). The pH value is usually rounded to one decimal place.
Calculate the pH of a solution with a hydrogen ion concentration of 2.5 × 10^-5 mol/L!
Answer:
For the solution we are calculating the pH of, the hydrogen ion concentration t = 2.5 × 10-5 is known, so we get:
The pH of the solution with a hydrogen ion concentration of t = 2.5 × 10-5 mol/L is -log(2.5 × 10-5) = -(-4 × 6) = 4×6. Therefore, the pH value of the solution is 4×6
9. Tyas is a student who likes to save money. Currently, Tyas’ savings have reached Rp5 million, but are only kept in a closet. Tyas then decides to deposit her money in a bank deposit with 5 percent interest per year. How many years will it take for Tyas’ savings to double?
Answer:
To make it easier, let’s assume that:
- Mₒ: Initial Capital = Rp5,000,000
- t: Time spent saving
- Mₜ: Capital after saving doubles = 2 x Mₒ = Rp10,000,000
- i: Annual deposit interest rate = 5% = 0.05.
So, every year, Tyas’ savings will be increased by the amount of 0.05 x 5000,000 = 250,000 in the first year, then 0.05 x 5,250,000 = 262,500 in the second year, and so on.
Therefore, the solution to the problem above can use the formula Mₜ = Mₒ x (1 + i)ᵗ.
Solution:
- Mₜ = Mₒ x (1 + i)ᵗ
- 10,000,000 = 5,000,000 x (1 + 0.05)ᵗ
- 1 + 0.05ᵗ = 10,000,000 / 5,000,000
- 1.05ᵗ = 2
Then, you can apply the logarithmic property log pⁿ = n x log p.
- log (1.05)ᵗ = log 2
- t x log (1.05) = log 2
- t = log 2 / (log 1.05)
- t = 0.3010 / 0.02119
- t = 14.21
In conclusion, Tyas’ savings will double after 14.2 years.
10.
Pengenalan Logaritma
Logaritma adalah salah satu konsep matematika yang penting dalam pemecahan masalah terkait perhitungan eksponensial. Logaritma digunakan untuk membalik operasi eksponensial dan membantu dalam menyelesaikan berbagai masalah matematika, fisika, dan sains lainnya. Dalam artikel ini, kita akan membahas teori dasar logaritma dan mengeksplorasi beberapa contoh soal yang dapat membantu memahami konsep ini dengan lebih baik.
Teori Dasar Logaritma
Logaritma didefinisikan sebagai kebalikan dari operasi eksponensial. Jika kita memiliki persamaan eksponensial \(a^b = c\), maka dalam bentuk logaritma akan menjadi \(\log_a c = b\). Di sini, \(a\) disebut sebagai basis logaritma, \(b\) adalah eksponen, dan \(c\) adalah hasil dari operasi eksponensial.
Beberapa properti logaritma yang penting untuk dipahami antara lain:
- \(\log_a (x \times y) = \log_a x + \log_a y\)
- \(\log_a (x / y) = \log_a x – \log_a y\)
- \(\log_a (x^b) = b \times \log_a x\)
Contoh Soal Logaritma
Contoh Soal 1
Tentukan nilai logaritma dari \(1/3 \log 6 + 1/2 \log 6\).
Jawaban:
\(1/3 \log 6 + 1/2 \log 6 = 6 \log 6 = 6\)
Jadi, nilai logaritma dari \(1/3 \log 6 + 1/2 \log 6\) adalah 6.
Contoh Soal 2
Hitung untuk menentukan nilai logaritma dari \(8 \log 32\).
Jawaban:
- \(8 \log 32 = 2^3 \log 2^5 = 5/3 \times 2 \log 2 = 5/3 \times 1 = 5/3\)
Kesimpulannya, nilai logaritma dari \(8 \log 32\) adalah 5/3 atau 1,6.
Contoh Soal 3
Hitung untuk menentukan nilai logaritma dari \(3 \log 18 – 3 \log 2\).
Jawaban:
- \(3 \log 18 – 3 \log 2 = 3 \log 18/2 = 3 \log 9 = 3 \log 3^2 = 2 \times 3 \log 3 = 2 \times 1 = 2\)
Contoh Soal 4
Sederhanakan bentuk logaritma \(2 \log 98 + 2 \log 50\).
Jawaban:
- \(2 \log 98 + 2 \log 50 = 2 \log (98 \times 50) = 2 \log 4900 = 2 \log (2^2 \times 5^2 \times 7^2) = 2 \times (2 \log 2 + 2 \log 5 + 2 \log 7) = 2 \times (1 + 2 + 2.322 + 2.808) = 2 \times 6.13 = 12.26\)
Contoh Soal 5
Sederhanakan bentuk logaritma \(3 \log 75 – 3 \log 6 + 9 + 2 \times 3 \log (1/9) + 3 \log (2 \frac{1}{4})\).
Jawaban:
- \(3 \log 75 – 3 \log 6 + 9 + 2 \times 3 \log (1/9) + 3 \log (2 \frac{1}{4})\)
- = \(3 \log (75 \times \frac{1}{81} \times \frac{9}{4}) / 6\)
- = \(3 \log \frac{25}{12} / 6\)
- = \(3 \log \frac{25}{12} / \frac{3}{2}\)
- = \(3 \log \frac{25}{12} \times \frac{2}{3}\)
- = \(3 \log \frac{50}{36}\)
- = \(3 \log \frac{25}{18}\)
- = 0.298
Contoh Soal 6
Jika diketahui \(3 \log 4 = a\) dan \(3 \log 5 = b\), tentukan \(8 \log 20\).
Jawaban:
- \(8 \log 20 = 3 \log 20 / 3 \log 8 = 3 \log (4 \times 5) / 3 \log (4 \times 2) = (3 \log 4 + 3 \log 5) / (3 \log 4 + 3 \log 2) = (a + b) / (a + a/2) = 2a + 2b / 3a\)
Contoh Soal 7
Hitung untuk menentukan nilai dari \((5 \log 10)^2 – (5 \log 2)^2 / 5 \log \sqrt{20}\).
Jawaban:
- \((5 \log 10)^2 – (5 \log 2)^2 / 5 \log \sqrt{20} = (5 \log 10 + 5 \log 2)(5 \log 10 – 5 \log 2) / 5 \log \sqrt{20} = (5 \log 20)(5 \log 5) / (1/2 \times 5 \log 20) = 1/1/2 = 2\)
Contoh Soal 8
Cari nilai logaritma dari \(4 \log 3\) jika diketahui \(9 \log 8 = 3m\).
Jawaban:
- \(3^2 \log 2^3 = 3m\)
- \(3/2 \times 3 \log 2 = 3m\)
- \(3 \log 2 = 2/3 \times 3m = 2m\)
Jadi, nilai dari \(4 \log 3\) adalah \(1/4m\).
Contoh Soal 9
Tentukan hasil nilai logaritma dari \(1/(3 \log 15) + (1/5 \log 15)\).
Jawaban:
- \(1/(3 \log 15) + (1/5 \log 15) = 15 \log 3 + 15 \log 5 = 15 \log (3 \times 5) = 15 \log 15 = 1\)
Jadi, hasil nilai logaritma dari \(1/(3 \log 15) + (1/5 \log 15)\) adalah 1.
Contoh Soal 10
Diketahui bahwa \(a \log b = 2\) DAN \(c \log b = 3\). Tentukan nilai logaritma dari \([a \log (bc)^3]^{1/3}\).
Jawaban:
- \(a \log b = 2\) diubah menjadi \(a = 2/\log b\)
- \(c \log b = 3\) diubah menjadi \(c = 3/\log b\)
- \([a \log (bc)^3]^{1/3}\) bisa disederhanakan menjadi \([a \times (3 \log bc)]^{1/3}\)
- \([a \times (3 \log bc)]^{1/3}\) dapat diubah menjadi \([a \times (3 \log b + \log c)]^{1/3}\)
- Jika nilai \(a\) dan \(c\) dimasukkan, ia jadi \(\{2/\log b \times [3 \times (\log b + 3/\log b)]\}^{1/3}\)
- Dengan penggabungan kurung, bentuk di atas menjadi:
- = \{6/\log b \times [(log b^2 + 3) / \log b]\}^{1/3}
- = \{[6 \times (\log b)^2 + 3] / (\log b)^2\}^{1/3}
- = \{6 \times [(\log b)^2 + 3 / (\log b)^2]\}^{1/3}
- Untuk cari hasil akhir, diasumsikan \(\log b = 1\), sehingga penyelesaiannya adalah:
- = \{6 \times [(1)^2 + 3 / (1)^2]\}^{1/3}
- = (6 \times 4)^{1/3}
- = 24^{1/3}
- = 2.884
Dengan demikian, karena \(a \log b = 2\) dan \(c \log b = 3\), nilai logaritma dari \([a \log (bc)^3]^{1/3}\) adalah 2.884.
Contoh Soal 11
Hitung untuk cari nilai logaritma dari \(3 \log 9 + 3 \log 18 + 3 \log 2\).
Jawaban:
- \(3 \log 9 + 3 \log 18 + 3 \log 2 = 3 \log (9 \times 18 \times 2) = 3 \log 324 = 3 \log 3^4 = 4 \times 3 \log 3 = 4 \times 1 = 4\)
Dapat disimpulkan, nilai dari \(3 \log 9 + 3 \log 18 + 3 \log 2\) adalah 4.
Dengan menjelajahi berbagai contoh soal logaritma, diharapkan pembaca dapat memahami konsep ini dengan lebih baik dan dapat mengaplikasikannya dalam pemecahan masalah matematika yang lebih kompleks. Jangan ragu untuk terus berlatih dan mendalami konsep logaritma agar kemampuan matematika Anda semakin terasah!
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