How do you solve $3^{2 x - 3} = 27 \sqrt{3}$ $3^(2x-3) = 27sqrt3$ ?

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Answer 1 · 2015-12-06T20:19:36

$x = 3, 25$ $x=3,25$

Using laws of exponents we may write this equation as

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

$therefore2x-3=3+1/2$

$thereforex=(6,5)/2=3,25$

How do you solve 3^(2x-3) = 27sqrt332x−3=27√33^(2x-3) = 27sqrt3?