- #1
James889
- 192
- 1
I have [tex]2lnx = xln2[/tex]
where [tex]x\ne2[/tex]
if you start by dividing both sides by ln2
is the following legal?
[tex]\frac{2lnx}{ln2} \rightarrow x = 2ln(x-2)[/tex]
[tex]e^{2ln(x-2)} = (x-2)^2[/tex]
[tex]x = (x-2)^2 \implies x = 4[/tex]
where [tex]x\ne2[/tex]
if you start by dividing both sides by ln2
is the following legal?
[tex]\frac{2lnx}{ln2} \rightarrow x = 2ln(x-2)[/tex]
[tex]e^{2ln(x-2)} = (x-2)^2[/tex]
[tex]x = (x-2)^2 \implies x = 4[/tex]