How do you solve $log x + ln x = 1$ $Logx+lnx=1$ ?

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How do you solve $log x + ln x = 1$ $Logx+lnx=1$ ?

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Answer 1 · 2016-08-28T13:40:04

$x = 2.00814$ $x = 2.00814$

Explanation:

${log}_{10} x = \frac{{log}_{e} x}{{log}_{e} 10}$ $log_(10)x = log_e x/(log_e 10)$

so

$\frac{{log}_{e} x}{{log}_{e} 10} + {log}_{e} x = 1$ $log_e x/(log_e 10)+log_e x = 1$

then

${log}_{e} x (\frac{1}{{log}_{e} 10} + 1) = 1$ $log_e x(1/log_e 10+1)=1$

${log}_{e} x = \frac{{log}_{e} 10}{1 + {log}_{e} 10}$ $log_e x = log_e 10/(1+log_e 10)$

$x = e^{\frac{{log}_{e} 10}{1 + {log}_{e} 10}} = 10^{\frac{1}{1 + {log}_{e} 10}} = 2.00814$ $x = e^( log_e 10/(1+log_e 10)) = 10^(1/(1+log_e 10)) = 2.00814$

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How do you solve $log x + ln x = 1$ $Logx+lnx=1$ ?

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Explanation:

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