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受欢迎的代数 >

x/(log_{332)(x)+1}=15

解答

\frac{x}{lo g _{332} ( x ) + 1} = 15

解答

x = - \frac{15 W _{- 1} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}, x = - \frac{15 W _{0} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}

+1

十进制

x = 23.11471 \dots, x = 0.00301 \dots

求解步骤

\frac{x}{lo g _{332} ( x ) + 1} = 15

在两边乘以

lo g_{332} (x) + 1

x = 15 (lo g_{332} (x) + 1)

交换两边

15 (lo g_{332} (x) + 1) = x

两边除以

15

lo g_{332} (x) + 1 = \frac{x}{15}

将

1

到右边

lo g_{332} (x) = \frac{x}{15} - 1

若

f (x) = g (x)

，则

a^{f (x)} = a^{g (x)}

33 2^{l o g_{332} (x)} = 33 2^{\frac{x}{15} - 1}

化简

33 2^{l o g_{332} (x)} : x

x = 33 2^{\frac{x}{15} - 1}

解

x = 33 2^{\frac{x}{15} - 1} : x = - \frac{15 W _{- 1} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}, x = - \frac{15 W _{0} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}

x = - \frac{15 W _{- 1} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}, x = - \frac{15 W _{0} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}

验证解:

x = - \frac{15 W _{- 1} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}

真

, x = - \frac{15 W _{0} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}

真

解为

x = - \frac{15 W _{- 1} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}, x = - \frac{15 W _{0} ( - \frac{l n ( 332 )}{4980} )}{ln ( 332 )}

作图

流行的例子

log_{10}(81x)= 1/4

lo g_{10} (81 x) = \frac{1}{4}

log_{10}(x)+log_{10}(x-21)=2

lo g_{10} (x) + lo g_{10} (x - 21) = 2

log_{3}(2x+1)+log_{3}(x+8)=3

lo g_{3} (2 x + 1) + lo g_{3} (x + 8) = 3

ln (x - 1) = 1

4*log_{10}(x)=3*(-0.6655)

4 \cdot lo g_{10} (x) = 3 \cdot (- 0.6655)