Manipulation Of Indices

Algebra Level 1

4 x = 2 3 y \large{4^x= \sqrt{2^{3y}}} Solve for x x from the equation above.

3 y 4 \dfrac{3y}{4} 3 y 3y y 3 \dfrac{y}{3} 4 y 3 \dfrac{4y}{3}

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4 solutions

Chew-Seong Cheong
Dec 16, 2015

4 x = 2 3 y 2 2 x = 2 3 y 2 2 x = 3 y 2 x = 3 y 4 \begin{aligned} 4^x & = \sqrt{2^{3y}} \\ 2^{2x} & = 2^{\frac{3y}{2}} \\ \Rightarrow 2x & = \frac{3y}{2} \\ x & = \boxed{\dfrac{3y}{4}} \end{aligned}

Sandeep Bhardwaj
Dec 16, 2015

The given equation is 4 x = 2 3 y 4^x=\sqrt{2^{3y}} which can be further written as 4 x = ( 2 3 y ) 1 / 2 4^x=\left( 2^{3y}\right)^{1/2} . Using the power rule , the equation becomes 4 x = 2 3 y / 2 4^x=2^{3y/2} . To make the bases equal we can write the equation as 2 2 x = 2 3 y / 2 2^{2x}=2^{3y/2} . Now as we have the bases equal, the exponents too must be equal for the equation to be correct. Hence we have 2 x = 3 y 2 x = 3 y 4 . 2x=\frac{3y}{2} \Rightarrow x=\frac{3y}{4}. \square

Moderator note:

Standard approach of simplifying the exponents into a common term.

Munem Shahriar
Sep 17, 2017

4 x = 2 3 y \large {4^x= \sqrt{2^{3y}}}

4 x = ( 2 3 y ) 1 2 \large 4^x = \left(2^{3y}\right)^{\frac{1}{2}}

4 x = ( 4 ) 3 y 2 \large 4^x = \left(\sqrt{4}\right)^{\frac{3y}{2}}

4 x = ( 4 1 2 ) 3 y 2 \large 4^x = \left(4^{\frac{1}{2}}\right)^{\frac{3y}{2}}

4 x = 4 3 y 2 × 1 2 \large 4^x = 4^{\frac{3y}{2} \times \frac{1}{2}}

4 x = 4 3 y 4 \large 4^x = 4^{\frac{3y}{4}}

x = 3 y 4 \large x = \boxed{\dfrac{3y}{4}}

4^x = √(2^[3y])

4^(2x) = 2^(3y)

2^(4x) = 2^(3y)

The same base

4x = 3y

x = 3y / 4

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