# A star 56 pc away has an apparent magnitude of 6.9. What is its absolute magnitude?

To solve, apply the formula:

`d=10*10^((m-M)/5)`

where d is the distance of the star from the Earth in parsecs

m is the apparent magnitude of the star, and

M is the absolute magnitude.

Plugging-in the values of d and m to the formula, it yields an equation:

`56=10 * 10^((6.9-M)/5)`

...

To solve, apply the formula:

`d=10*10^((m-M)/5)`

where
d is the distance of the star from the Earth in parsecs

m is the apparent magnitude of the star, and

M is the absolute magnitude.

Plugging-in the values of d and m to the formula, it yields an equation:

`56=10 * 10^((6.9-M)/5)`

Then, solve for M. To do so, divide both sides by 10.

`5.6=10^((6.9-M)/5)`

Then, take the logarithm of both sides.

`log5.6=log(10^((6.9-M)/5))`

The right side simplifies to:

`log5.6=(6.9-M)/5`

Then, multiply both sides by 5.

`5log5.6=6.9-M`

Subtract 6.9 from both sides.

`5log5.6 - 6.9=-M`

And, divide both sides by -1.

`-5log5.6 + 6.9=M`

`6.9- 5log5.6=M`

`3.159059864=M`

Rounding off to nearest tenths, it becomes:

`3.2=M`

Therefore, the absolute magnitude of the star is 3.2 .

Approved by eNotes Editorial Team