How do I solve lnx+ ln(2x+1)=0

Question

anonymous · Answer

the answer is 1/2 but I don't know how to get there

jim_thompson5910 · Answer

ln(x)+ ln(2x+1)=0

ln(x*(2x+1))=0

x*(2x+1) = e^0

x*(2x+1) = 1

2x^2 + x = 1

2x^2 + x - 1 = 0

I'll let you finish. Don't forget to check your answers.

anonymous · Answer

I solved, got the answer, and understand everything except the RHS of the third step...how do I know that 0 is supposed to be e^0 and that e^0 means one? or what does e mean?

jim_thompson5910 · Answer

I'm using the idea that if 

ln(x) = y

then

x = e^y

jim_thompson5910 · Answer

since in general

$$\large \log_{b}(x) = y$$

turns into

$$\large x = b^y$$

jdoe0001 · Answer

$\bf ln(x \times (2x+1))=0 \implies log_e(x \times (2x+1))=0\
\textit{exponentialize both sides by "e"}\
 \color{blue}{\Large e^{log_e(x \times (2x+1))}  = e^{0}}\
\textit{log cancellation rule of } \Large a^{log_ax} = x\
x \times (2x+1) = e^{0}$

anonymous · Answer

Thanks to both of you!

jim_thompson5910 · Answer

yw