How do you solve $(2x+7)^(3/2)=27$ ?

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Answer 1 · 2016-03-19T15:32:05

We must first convert to radical form by using the exponent rule $a^(m/n) = root(n)(a^m)$

$sqrt((2x + 7)^3) = 27$

We can get rid of the radical by squaring both sides of the equation.

$(2x + 7)^3 = 729$

Now, to cancel the cube, we must take the cube root of $729$ .

$2x + 7 = 9$

$2x = 2$

$x = 2/2$

$x = 1$

Practice exercises:

Solve for x:

a). $(2x + 1)^(3/2)= 7x - 1$

b). $8^(x/3) = 3x + 4$

Good luck!

How do you solve (2x+7)^(3/2)=27(2x+7)^(3/2)=27?