How do you solve #3x ^ { \frac { 3} { 4} } = x ^ { \frac { 1} { 2} }#?

1 Answer
smendyka
Sep 10, 2017

See a solution process below:

Explanation:

First, divide each side of the equation by #color(blue)(3)color(red)(x^(1/2))# to isolate the #x# term while keeping the equation balanced:

#(3x^(3/4))/(color(blue)(3)color(red)(x^(1/2))) = x^(1/2)/(color(blue)(3)color(red)(x^(1/2)))#

#(color(blue)(cancel(color(black)(3)))x^(3/4))/(cancel(color(blue)(3))color(red)(x^(1/2))) = color(red)(cancel(color(black)(x^(1/2))))/(color(blue)(3)cancel(color(red)(x^(1/2))))#

#x^(3/4)/x^(1/2) = 1/3#

Now, use this rule of exponents to simplify the expression on the left side of the equation:

#x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#x^color(red)(3/4)/x^color(blue)(1/2) = 1/3#

#x^(color(red)(3/4)-color(blue)(1/2)) = 1/3#

#x^(color(red)(3/4)-(2/2 xx color(blue)(1/2))) = 1/3#

#x^(color(red)(3/4)-color(blue)(2/4)) = 1/3#

#x^(1/4) = 1/3#

Now, put both sides of the equation to the fourth power to solve for #x# using these rules of exponents:

#(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))# and #a^color(red)(1) = a#

#(x^color(red)(1/4))^color(blue)(4) = (1/3)^4#

#x^(color(red)(1/4) xx color(blue)(4)) = 1/81#

#x^(4/4) = 1/81#

#x^1 = 1/81#

#x = 1/81#

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