Answer:
[tex]\begin{aligned}\textsf{(a)}\quad &150 = 2 \times 3 \times 5^2\\&\log_3(150)\approx 4.561\end{aligned}[/tex]
[tex]\textsf{(b)}\quad \log_{3}(150)=\dfrac{\log(150)}{\log(3)}=4.561\; \sf (nearest\;thousandth)[/tex]
Step-by-step explanation:
The prime factorization of a number involves expressing it as the product of its prime factors. Therefore, the prime factorization of 150 is:
[tex]150 = 2 \times 3 \times 5 \times 5[/tex]
[tex]150 = 2 \times 3 \times 5^2[/tex]
To evaluate log₃(150) given the approximations log₃(2) ≈ 0.631 and log₃(5) ≈ 1.465, we can rewrite 150 as a product of its prime factors, and then apply the properties of logarithms.
[tex]\boxed{\begin{array}{c}\underline{\sf Logarithmic\;Properties}\\\\\textsf{Product:}\quad \log_axy=\log_ax + \log_ay\\\\\textsf{Power:}\quad\log_ax^n=n\log_ax\qquad\;\end{array}}[/tex]
Begin by rewriting 150 as the product of its prime factors:
[tex]\log_3(150)=\log_3(2\cdot 3 \cdot 5^2)[/tex]
Apply the product property of logarithms:
[tex]=\log_3(2)+\log_3(3)+\log_3(5^2)[/tex]
Now, apply the power property of logarithms:
[tex]=\log_3(2)+\log_3(3)+2\log_3(5)[/tex]
Given that logₐ(a) = 1, then:
[tex]=\log_3(2)+1+2\log_3(5)[/tex]
Finally, substitute the given approximations for log₃(2) and log₃(5):
[tex]=0.631+1+2(1.465)\\\\=0.631+1+2.93\\\\=4.561[/tex]
Therefore, the evaluation of log₃(150) is 4.561, rounded to the nearest thousandth.
The change of base formula is:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Change of base formula}}\\\\\log_ba=\dfrac{\log_xa}{\log_xb}\end{array}}[/tex]
When using the LOG function on most calculators, it typically computes logarithms with base 10. Therefore, to evaluate log₃(150) using the calculator's LOG function, we need to change the base to 10 using the change of base formula:
[tex]\log_{3}(150)=\dfrac{\log_{10}(150)}{\log_{10}(3)}[/tex]
When we write the log of a number without specifying a base, it is understood to be the logarithm with base 10. Therefore:
[tex]\log_{3}(150)=\dfrac{\log(150)}{\log(3)}[/tex]
Using a calculator to compute this gives:
[tex]\begin{aligned}\log_{3}(150)&=\dfrac{\log(150)}{\log(3)}\\\\&=\dfrac{2.176091259...}{0.477121254...}\\\\&=4.560876795...\\\\&=4.561\; \sf (nearest\;thousandth)\end{aligned}[/tex]
To find the prime factorization of 150, divide it by the smallest prime numbers until you cannot divide further. Then, using the properties of logarithms and the given approximations, evaluate log3 150.
Given log3 2 ≈ 0.631 and log3 5 ≈ 1.465
(a) To find the prime factorization of 150, we start by dividing it by the smallest prime number, which is 2. 150 divided by 2 is 75. We can continue dividing 75 by 2, and we get 37.5, which is not a whole number. So, we move to the next prime number, which is 3. 75 divided by 3 is 25, and 25 divided by 5 is 5. Therefore, the prime factorization of 150 is 2 x 3 x 5 x 5.
To evaluate log3 150, we can use the properties of logarithms. Using the property that logb (mn) = logb (m) + logb (n), we can rewrite 150 as (2 x 3 x 5 x 5). Then, using the approximation log3 2 ≈ 0.631 and log3 5 ≈ 1.465, we can evaluate each term. log3 (2 x 3 x 5 x 5) = log3 2 + log3 3 + log3 5 + log3 5 ≈ 0.631 + 1 + 1.465 + 1.465 ≈ 4.561.
(b) To evaluate log3
150 using the change of base formula, we can use the fact that loga
b = logc
b / logc
a. Let's use the common logarithm as the base, which is log10. log3
150 = log10
150 / log10
3. Using a calculator, log10
150 ≈ 2.176, and log10
3 ≈ 0.477. Therefore, log3
150 ≈ 2.176 / 0.477 ≈ 4.561 (rounded to the nearest thousandth).
