Answer:
Option 1 - [tex]\log_a (3c-d)=2b+1[/tex]
Step-by-step explanation:
Given : Equation [tex]a^{2b+1}=3c-d[/tex]
To find : Which of the following is the equation ?
Solution :
Equation [tex]a^{2b+1}=3c-d[/tex]
Taking log both side,
[tex]\log (a^{2b+1})=\log (3c-d)[/tex]
Apply logarithmic property, [tex]\log a^x=x\log a[/tex]
[tex](2b+1)\log (a)=\log (3c-d)[/tex]
[tex](2b+1)=\frac{\log (3c-d)}{\log a}[/tex]
Apply change the base rule, [tex]\frac{\log _a x}{\log _a y}=\log_y x[/tex]
[tex](2b+1)=\log_a (3c-d)[/tex]
So, the required equation is [tex]\log_a (3c-d)=2b+1[/tex]
Therefore, option 1 is correct.
