Question 1
Solve the following equation for y:
log327y + 1 = log352y
- 3⁄log325 - 3
- log35 - 1
- 2
- 27-Jan
- log35
Explanation: log327 y + 1 = log35 2y Rewrite 27 as 33.
log33 3(y + 1) = log352y
3(y + 1) = 2y · log35
3y + 3 = 2y · log35
3⁄log325 - 3
log33 3(y + 1) = log352y
3(y + 1) = 2y · log35
3y + 3 = 2y · log35
3⁄log325 - 3