Question

Pls how do I solve
b) log₂ (7y-1)=3+log₂ (y-1).

Answer 1

Explanation:

Put the logs on the same side.

`log_(2)(7y - 1) - log_(2)(y - 1) = 3`

Using log rules, whenever you separate two logs using subtraction and they have the same base, we can write them as one log where they are a rational function

`log_(2)( (7y - 1)/(y - 1) ) = 3`

Let both of these expression be the power of a exponent with base 2.

`2 {}^{ log_(2)( (7y - 1)/(y - 1) ) } = 2 {}^(3)`

The base 2 and log base 2 cancel out

`(7y - 1)/(y - 1) = 8`

When y =1, we will get undefined so y cannot be 1

However, multiply both sides by y-1

`7y - 1 = 8(y - 1)`

`7y - 1 = 8y - 8`

`7 = y`