Answer:
a) u - v
b) 2v - 2
c) 3u + 3
d) ¹/₂
Step-by-step explanation:
Given:
[tex]u=\log_{10}3[/tex]
[tex]v=\log_{10}5[/tex]
Part (a)
Rewrite 0.6 as a fraction:
[tex]\implies \log_{10}(0.6)=\log_{10}\left(\dfrac{3}{5}\right)[/tex]
[tex]\textsf{Apply the quotient log law}: \quad \log_a\frac{x}{y}=\log_ax - \log_ay:[/tex]
[tex]\implies \log_{10}\left(\dfrac{3}{5}\right)=\log_{10}3-\log_{10}5[/tex]
Substitute the values of u and v:
[tex]\implies \log_{10}3-\log_{10}5=u-v[/tex]
Part (b)
Rewrite 0.25 as 25/100:
[tex]\implies \log_{10}(0.25)=\log_{10}\left(\dfrac{25}{100}\right)[/tex]
[tex]\textsf{Apply the quotient log law}: \quad \log_a\frac{x}{y}=\log_ax - \log_ay[/tex]
[tex]\implies \log_{10}\left(\dfrac{25}{100}\right)=\log_{10}(25)-\log_{10}(100)[/tex]
Rewrite 25 as 5² and 100 as 10²:
[tex]\implies \log_{10}(25)-\log_{10}(100)=\log_{10}(5^2)-\log_{10}(10^2)[/tex]
[tex]\textsf{Appy the Power log law}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies \log_{10}(5^2)-\log_{10}(10^2)=2\log_{10}5-2\log_{10}10[/tex]
[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]
[tex]\implies 2\log_{10}5-2\log_{10}10=2\log_{10}5-2(1)[/tex]
Substitute the value of v:
[tex]\implies 2\log_{10}5-2(1)=2v-2[/tex]
Part (c)
Rewrite 27000 as 30³:
[tex]\implies \log_{10}(27000)=\log_{10}(30^3)[/tex]
[tex]\textsf{Appy the Power log law}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies \log_{10}(30^3)=3\log_{10}(30)[/tex]
[tex]\textsf{Apply the log product law}: \quad \log_axy=\log_ax + \log_ay[/tex]
[tex]\implies 3\log_{10}(30)=3\log_{10}(3)+3\log_{10}(10)[/tex]
[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]
[tex]\implies 3\log_{10}(3)+3\log_{10}(10)=3\log_{10}(3)+3(1)[/tex]
Substitute the value of u:
[tex]\implies 3\log_{10}(3)+3(1)=3u+3[/tex]
Part (d)
Rewrite √10 as [tex]10^{\frac{1}{2}}[/tex] :
[tex]\implies \log_{10}(\sqrt{10})=\log_{10}(10^{\frac{1}{2}})[/tex]
[tex]\textsf{Appy the Power log law}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies \log_{10}(10^{\frac{1}{2}})=\dfrac{1}{2}\log_{10}(10)[/tex]
[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]
[tex]\implies \dfrac{1}{2}\log_{10}(10)=\dfrac{1}{2}(1)=\dfrac{1}{2}[/tex]
