Question

Rewrite the given expression in the form 3^u where u is a constant or an algebraic expression.

Answer 1

Simplifying the given expressions we proceed as follows:
(5sqrt3)^x
=5^x*(3^1/2)^x
=5^x*3^x/2
=5^x3^u
where u=x/2


(1/2)^(x-3)
=1/2^(x-3)
=2^-(3-x)
=2^u
where u=-(3-x)

9/3sqrt(3)
=3/(3)^(1/2)
=3(3)^(-1/2)

16/(3sqrt (2^x))
=1/3*(2^4*2^(-x/2))
=1/3*2^(4-x/2)
=1/3*2^u
where:
u=4-x/2

Answer 2

Solution: (1) The expression $(5\sqrt{3} )^x$  $\text{ is written as }$ $5^x3^{\frac{x}{2}$.

$(5\sqrt{3} )^x=5^x(\sqrt{3} )^x\\(5\sqrt{3} )^x=5^x(3^\frac{1}{2} )^x\\(5\sqrt{3} )^x=5^x3^\frac{x}{2}$

The value of u is $\frac{x}{2}$.

(2) The expression $(\frac{1}{2})^x-3$ is written as $2^{-x}-3$.

$(\frac{1}{2})^x-3=(2^{-1})^x-3\\(\frac{1}{2})^x-3=2^{-x}-3$

The value of u is -x.

(3) The expression $\frac{9}{3\sqrt{3} }$ is written as $\sqrt{3} (2^0)$.

$\frac{9}{3\sqrt{3} }=\frac{9}{3\sqrt{3} }(\frac{\sqrt{3}}{\sqrt{3}} )\\\frac{9}{3\sqrt{3} }=\frac{9(\sqrt{3}) }{3(3) }\\\frac{9}{3\sqrt{3} }=\sqrt{3} \\\frac{9}{3\sqrt{3} }=\sqrt{3}(2^0)$

THe value of u is 0.

(4)The expression $\frac{16}{3(\sqrt{2^{x}} )}$ is written as $\frac{1}{3}(2^{4-\frac{x}{2}})$.

$\frac{16}{3(\sqrt{2^{x}} )}=\frac{2^4}{3(2^x)^{1/2}}\\\frac{16}{3(\sqrt{2^{x}} )}=\frac{1}{3}(2^{4-\frac{x}{2}})$

The value of u is $4-\frac{x}{2}$.