Answer and Step-by-step explanation:
Part A: Exponential equation is the "opposite" of logarithmic equation, so:
[tex]y=log_{b}m[/tex]
[tex]b^{y}=m[/tex]
Part B: Using log with base c:
[tex]log_{c}b^{y}=log_{c}m[/tex]
Power property of logarithm states that if the anti-logarithm is elevated at a power, the elevated number can be pulled in front of the logarithm:
[tex]ylog_{c}b=log_{c}m[/tex]
Solving for y:
[tex]y=\frac{log_{c}m}{log_{c}b}[/tex]
Part C: To facilitate the use of calculators, which only have values for the base-10 log and natural log, we use change of base formula, i.e., transform
[tex]y=log_{b}m[/tex]
into
[tex]y=\frac{log_{c}m}{log_{c}b}[/tex]
Part D: [tex](log_{3}z)(log_{z}27)[/tex]
Change of base will be:
[tex]log_{3}z=\frac{log_{10}z}{log_{10}3}[/tex]
[tex]log_{z}27=\frac{log_{10}27}{log_{10}z}[/tex]
Solving:
[tex](log_{3}z)(log_{z}27)[/tex] = [tex](\frac{log_{10}z}{log_{10}3})(\frac{log_{10}27}{log_{10}z} )[/tex]
[tex](log_{3}z)(log_{z}27)[/tex] = [tex]\frac{log_{10}27}{log_{10}3}[/tex]
[tex](log_{3}z)(log_{z}27)[/tex] = [tex]\frac{log_{10}3^{3}}{log_{10}3}[/tex]
[tex](log_{3}z)(log_{z}27)[/tex] = [tex]\frac{3log_{10}3}{log_{10}3}[/tex]
[tex](log_{3}z)(log_{z}27)[/tex] = 3
Part E: [tex]log_{7}300[/tex]
Using change of base:
[tex]log_{7}300=\frac{log300}{log7}[/tex]
[tex]log_{7}300=\frac{2.48}{0.85}[/tex]
[tex]log_{7}300[/tex] ≈ 3