Answer:
[tex]x=\frac{1}{4}[/tex]
Step-by-step explanation:
We can use some logarithmic rules to solve this easily.
Note: Ln means [tex]Log_e[/tex]
Now, lets start with the equation:
[tex]ln(2x) + ln(2) = 0\\ln(2x) = -ln(2)[/tex]
Writing left side with logarithmic base e, we have:
[tex]Log_{e}(2x) = -ln(2)[/tex]
We can now use the property shown below to make this into exponential form:
[tex]Log_{a}b=x\\means\\a^x=b[/tex]
So, we write:
[tex]Log_{e}(2x) = -ln(2)\\e^{-ln(2)}=2x[/tex]
We recognize another property of exponentials:
[tex]a^{bc}=(a^{b})^{c}[/tex]
So, we write:
[tex]e^{-ln(2)}=2x\\(e^{ln(2)})^{-1}=2x[/tex]
Also, another property of natural logarithms is:
[tex]e^{(ln(a))}=a[/tex]
Now, we simplify:
[tex](e^{ln(2)})^{-1}=2x\\(2)^{-1}=2x\\\frac{1}{2}=2x\\x=\frac{\frac{1}{2}}{2}\\x=\frac{1}{4}[/tex]
This is the answer.
