SOLUTION: my teacher did not get to explain these kinds of problems before the bell rang. in this equation, # and & is what we must find. a^#/3b^&=b^3/3 can you help explain this to m

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: my teacher did not get to explain these kinds of problems before the bell rang. in this equation, # and & is what we must find. a^#/3b^&=b^3/3 can you help explain this to m      Log On


   



Question 122371This question is from textbook prentice hall mathmatics algebra 1
: my teacher did not get to explain these kinds of problems before the bell rang.
in this equation, # and & is what we must find.
a^#/3b^&=b^3/3
can you help explain this to me?
i started this but i think i got it all wrong. thank you sooo much!
emma
titusville, fl
This question is from textbook prentice hall mathmatics algebra 1

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
there is no "a" term on the right-hand side of the equation
__ the only way this can happen is if #=0 (a^0=1)
__ if # had any other value, there would be an "a" term on both sides

so you have 1/3b^&=b^3/3

multiplying by 3 clears the fractions __ b^&=b^3

so &=3