Given:
[tex]x=\log_{10}31,500[/tex]
To find:
Select the true statements from the given options about the given value.
Solution:
We have,
[tex]x=\log_{10}31,500[/tex]
It can be written as
[tex]x=\log_{10}(2^23^25^37)[/tex]
[tex]x=\log_{10}(2^2)+\log_{10}(3^2)+\log_{10}(5^3)+\log_{10}(7)[/tex] [tex][\because \log(ab)=\log a+\log b][/tex]
[tex]x=2\log_{10}2+2\log_{10}3+3\log_{10}5+\log_{10}(7)[/tex]
[tex]x=2\left(0.30105\right)+2\left(0.47712\right)+3\left(0.69897\right)+0.8451[/tex]
[tex]x=4.49835[/tex]
Clearly, the value of x lies between 4 and 5. So, [tex]x>4[/tex] and [tex]x<5[/tex].
Therefore, the correct options are C and D.
The true statements are: (c) x > 4 and (d). x < 5
The logarithmic expression is given as:
[tex]x = \log_{10}(31500)[/tex]
Express 31500 as a product of 100
[tex]x = \log_{10}(315 * 100)[/tex]
Apply the law of logarithm
[tex]x = \log_{10}(315) + \log_{10}(100)[/tex]
Evaluate log 100 base 10
[tex]x = \log_{10}(315) + 2[/tex]
Evaluate log 315 base 10
[tex]x = 2.50 + 2[/tex]
Add the expressions
[tex]x \approx 4.50[/tex]
4.5 is greater than 4 and less than 5
Hence, the true statements are: (c) x > 4 and (d). x < 5
Read more about logarithms at:
https://brainly.com/question/25710806
