How do you solve #\sqrt { 125} = 5^ { x }#?

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Answer 1 · 2018-06-30T20:54:43

# x = 3/2 #

#sqrt(125) = 125^(1/2) #

#125 = 5^3 #

#=> 125^(1/2) = (5^3)^(1/2) #

#=> 5^x = (5^3 ) ^(1/2) #

#=> 5^x = 5^(3/2) #

Considering the exponents:

# x = 3/2 #

Answer 2 · 2018-06-30T21:08:23

#x=3/2#

We can rewrite #sqrt125# as #125^(1/2)#. This now gives us

#125^(1/2)=5^x#

Let's make our bases the same. #125=5^3#, so we can rewrite the equation as

#5^(3(1/2))=5^x#

Since our bases are the same, the exponents are equal.

#5^(3/2)=5^x#

#=>3/2=x#

#x=3/2#

Hope this helps!