Using logarithms to solve fractional exponents

Other day I was studying logarithms in a basic level math course that revisited almost everything we were supposed to learn at high school when I was shocked to see how one of its properties could be easily applied to solve fractional exponents. I confess that I felt thrilled when I saw it.

Basically the property used to solve it is the one that states that:

log₂8² = 2.log₂8

Just to check this equality, the solution of both sides give us:

 log₂8² = log₂64 = 6

to obtain 64, 2 must be raised to the exponent 6

2.log₂8 = 2.3 = 6

So in both cases we get the same result - the property is true!

Now what about this:

y = 25^1/2 ?

We know this is equivalent to the square root of 25, but lets solve it using logarithms

The equality signal (=) was originally a representation of a balance scale. Both sides have to have the same "weight" to keep the balance. So, whatever we do in one side, we also do in the other.

log y = log 25^1/2

Applying the property we get:

log y = 1/2 log 25

Let's now define the base of the logarithm. In this case, base 5 seems quite convenient:
(It is important to remember that the base we choose must be the same used by all logs in our equation)

log₅ y = 1/2 log₅ 25 

log₅ y = 1/2 . 2

log₅ y = 1

y = 5¹ 

y = 5

Now lets consider a more complete example using almost all properties of logarithms - which can be checked out here

Adding the log in both sides we get:

Now using the properties - when a number is divided we subtract the log of both and when a number is multiplied we add - our equation gets the form:

log h = log (81)^1/3 + log (27)^1/2.5 - log (243)^1/10

It is clear from the example that the numbers 81, 27 and 243 are all power of 3, so it will be chosen as the base.

log₃ h = log₃ (81)^1/3 + log₃ (27)^1/2.5 - log₃ (243)^1/10

log₃ h = 1/3 log₃ (81) + 1/2.5 log₃ (27) - 1/10 log₃ (243)

Reckoning the logs at the right side we get


log₃ h = 1/3 4 +  1/2.5 3 - 1/10 5
log₃ h = 1.33 + 1.2 - 0.5
log₃ ˜ 2

Reckoning the inverse of the logarithm, the exponentiation we get the final result - approximate

h ˜ 3²
h ˜ 9


Voilà, logarithms are really a powerful tool, thanks to John Napier!