[數學]2^(2x+5) - 4^(x+1)=448

2011-08-11 3:15 am

解指數方程2^(2x+5) - 4^(x+1)=448

唔記得哂點計,點算>3<
麻煩哂....

回答 (3)

2011-08-11 3:40 am

✔ 最佳答案

2^(2x+5)-4^(x+1)=448
(2^5)*(2^2x)-(4^1)*(4^x)=448
32(4^x)-4(4^x)=448
28(4^x)=448
4^x=16
4^x=4^2
x=2

參考: just my answer

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2011-08-11 7:03 am

2^(2x+5) - 4^(x+1)=448
(2^2x)(2^5) - (2^2x)(2^2)=448
2^2x(2^5-2^2)=448
2^2x=16
2^2x=2^4
2x=4
x=2

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2011-08-11 3:39 am

2^(2x+5) - 4^(x+1)=448
2^(2x+2) * 2^3 - 4^(x+1)=448
8* 4^(x+1) - 4^(x+1)=448
7* 4^(x+1)=448
4^(x+1)=64
4^(x+1)=4^3
x+1=3
x=2

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