Answer:
3.876 (3 d.p.)
Step-by-step explanation:
Given:
- [tex]\log_{10}2=0.3010[/tex]
- [tex]\log_{10}3=0.4771[/tex]
Please note: The question gives two values of log base 10 which should be used to help find the value of [tex]\log_5512[/tex]
Log Law: Change of Base
[tex]\log_ba=\dfrac{\log_xa}{\log_xb}[/tex]
Change the given expression to log base 10:
[tex]\implies \log_5512=\dfrac{\log_{10}512}{\log_{10}5}[/tex]
Replace 512 with 2⁹, and 5 with 10/2:
[tex]\implies \dfrac{\log_{10}{2^9}}{\log_{10}\frac{10}{2}}[/tex]
[tex]\textsf{Apply the Power Log law}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies \dfrac{9\log_{10}{2}}{\log_{10}\frac{10}{2}}[/tex]
[tex]\textsf{Apply the Log Quotient law}: \quad \log_a\frac{x}{y}=\log_ax - \log_ay[/tex]
[tex]\implies \dfrac{9\log_{10}{2}}{\log_{10}10-\log_{10}2}[/tex]
[tex]\textsf{Apply log law}: \quad \log_aa=1[/tex]
[tex]\implies \dfrac{9\log_{10}{2}}{1-\log_{10}2}[/tex]
Given:
- [tex]\log_{10}2=0.3010[/tex]
Substitute this into the expression and simplify:
[tex]\implies \dfrac{9(0.3010)}{1-0.3010}[/tex]
[tex]\implies \dfrac{2.709}{0.699}[/tex]
[tex]\implies 3.876\:(\sf 3\: d.p.)[/tex]
