Find the value of x; if:
(23)3×(23)−4=(23)2x+1\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^{-4} = \Big(\dfrac{2}{3}\Big)^{2x+1}(32)3×(32)−4=(32)2x+1
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(23)3×(23)−4=(23)2x+1⇒(23)3+(−4)=(23)2x+1⇒3−4=2x+1⇒−1=2x+1⇒−1−1=2x⇒−2=2x⇒x=−22⇒x=−1\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^{-4} = \Big(\dfrac{2}{3}\Big)^{2x+1}\\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{3+(-4)} = \Big(\dfrac{2}{3}\Big)^{2x+1}\\[1em] \Rightarrow 3 - 4 = 2x + 1\\[1em] \Rightarrow -1 = 2x + 1\\[1em] \Rightarrow -1 - 1 = 2x\\[1em] \Rightarrow -2 = 2x\\[1em] \Rightarrow x = \dfrac{-2}{2}\\[1em] \Rightarrow x = -1(32)3×(32)−4=(32)2x+1⇒(32)3+(−4)=(32)2x+1⇒3−4=2x+1⇒−1=2x+1⇒−1−1=2x⇒−2=2x⇒x=2−2⇒x=−1
If (23)3×(23)−4=(23)2x+1\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^{-4} = \Big(\dfrac{2}{3}\Big)^{2x+1}(32)3×(32)−4=(32)2x+1, then x=−1x = -1x=−1.
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Evaluate:
5n+2−5n+15n+1\dfrac{5^{n+2}-5^{n+1}}{5^{n+1}}5n+15n+2−5n+1
1(125)x−7=52x−1\dfrac{1}{(125)^{x-7}}=5^{2x-1}(125)x−71=52x−1
4n÷4−3=454^n ÷ 4^{-3} = 4^54n÷4−3=45
Simplify:81×3n+1−9×3n81×3n+2−9×3n+1\dfrac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}}81×3n+2−9×3n+181×3n+1−9×3n
