SOLUTION: Solve for the unknown variable: a)sqrt9^x=27 b)8^(1/2x)=(2^7)*(4^9) c)sqrt125^x=5/25^x Can someone PLEASE help me ???

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve for the unknown variable: a)sqrt9^x=27 b)8^(1/2x)=(2^7)*(4^9) c)sqrt125^x=5/25^x Can someone PLEASE help me ???      Log On


   

Question 125675: Solve for the unknown variable:
a)sqrt9^x=27
b)8^(1/2x)=(2^7)*(4^9)
c)sqrt125^x=5/25^x Can someone PLEASE help me ???
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
a) sqrt%289%5Ex%29=27
Square both sides:
9%5Ex=27%5E2 (Don't bother calculating 27%5E2, it won't be necessary)
But 9=3%5E2 and 27=3%5E3, so
%283%5E2%29%5Ex=%283%5E3%29%5E2

Use the rule: %28x%5Em%29%5En=x%5Emn
3%5E%282x%29=3%5E6

So, 2x=6
x=3

Check:
9%5E3=729
sqrt%28729%29=27, answer checks.

b) Do you mean: 8%5E%281%2F2x%29=%282%5E7%29%2A%284%5E9%29 or do you really mean: 8%5E%28%281%2F2%29x%29=%282%5E7%29%2A%284%5E9%29

Either way: Notice that 8=2%5E3 and 4=2%5E2, so:

Either:
2%5E%283%2F2x%29=%282%5E7%29%2A%282%5E18%29=2%5E25
or
2%5E%28%283%2F2%29x%29=%282%5E7%29%2A%282%5E18%29=2%5E25

So either 3%2F2x=25 or %283x%2F2%29=25
And x=3%2F50 or x=50%2F3

You should be able to use the same techniques to solve the last one. For part c get everything represented by powers of 5. 25=5%5E2 and 125=5%5E3