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
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81×3n+1−9×3n81×3n+2−9×3n+1=34×3n+1−32×3n34×3n+2−32×3n+1=34+(n+1)−32+n34+(n+2)−32+(n+1)=3n+5−32+n3n+6−3n+3=(3n+2.33)−32+n(3n+3.33)−3n+3=(3n+2)(3n+3)33−133−1=(3(n+2)−(n+3))33−133−1=(3n+2−n−3)=(3−1)=13\dfrac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}}\\[1em] = \dfrac{3^4\times3^{n+1}-3^2\times3^n}{3^4\times3^{n+2}-3^2\times3^{n+1}}\\[1em] = \dfrac{3^{4+(n+1)}-3^{2+n}}{3^{4+(n+2)}-3^{2+(n+1)}}\\[1em] = \dfrac{3^{n+5}-3^{2+n}}{3^{n+6}-3^{n+3}}\\[1em] = \dfrac{(3^{n+2}.3^3)-3^{2+n}}{(3^{n+3}.3^3)-3^{n+3}}\\[1em] = \dfrac{(3^{n+2})}{(3^{n+3})}\dfrac{3^3-1}{3^{3}-1}\\[1em] = (3^{(n+2)-(n+3)})\dfrac{\cancel {3^3-1}}{\cancel {3^3-1}}\\[1em] = (3^{n+2-n-3})\\[1em] = (3^{-1})\\[1em] = \dfrac{1}{3}81×3n+2−9×3n+181×3n+1−9×3n=34×3n+2−32×3n+134×3n+1−32×3n=34+(n+2)−32+(n+1)34+(n+1)−32+n=3n+6−3n+33n+5−32+n=(3n+3.33)−3n+3(3n+2.33)−32+n=(3n+3)(3n+2)33−133−1=(3(n+2)−(n+3))33−133−1=(3n+2−n−3)=(3−1)=31
81×3n+1−9×3n81×3n+2−9×3n+1=13\dfrac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}} = \dfrac{1}{3}81×3n+2−9×3n+181×3n+1−9×3n=31
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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
4n÷4−3=454^n ÷ 4^{-3} = 4^54n÷4−3=45
If 2n−7×5n−4=12502^{n-7} \times 5^{n-4} = 12502n−7×5n−4=1250, find n.
The multiplicative inverse of (80+50)(80−50)(8^0 + 5^0)(8^0 - 5^0)(80+50)(80−50) is:
0
49
1
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