eln⁡2+813+lg⁡20−lg⁡2=e^{\ln 2}{+}8^{\frac{1}{3}}{+}\lg 20{-}\lg 2{=}___

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Worked solution

1. 计算指数部分的值

根据对数的性质 alog⁡ax=xa^{\log_a x} = x 可得:
eln⁡2=2e^{\ln 2} = 2
根据指数幂的运算法则可得:
813=(23)13=28^{\frac{1}{3}} = (2^3)^{\frac{1}{3}} = 2

2. 计算对数部分的值

根据对数的运算性质 lg⁡A−lg⁡B=lg⁡AB\lg A - \lg B = \lg\frac{A}{B} 可得:
lg⁡20−lg⁡2=lg⁡202=lg⁡10=1\lg 20 - \lg 2 = \lg \frac{20}{2} = \lg 10 = 1

3. 求和得出最终结果

将各部分化简后的值相加:
2+2+1=52 + 2 + 1 = 5

Answer

5

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