解不等式log2(4^x+1)>log(2^x+7)
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解不等式log2(4^x+1)>log(2^x+7)
![解不等式log2(4^x+1)>log(2^x+7)](/uploads/image/z/3938413-13-3.jpg?t=%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Flog2%284%5Ex%2B1%29%3Elog%282%5Ex%2B7%29)
log2(4^x+1)>log(2^x+7)
所以
2(4^x+1)>(2^x+7)
2^(2x+2)>2^(x+7)
即
2x+2>x+7
x>7-2
x>5
所以
2(4^x+1)>(2^x+7)
2^(2x+2)>2^(x+7)
即
2x+2>x+7
x>7-2
x>5
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