\(e ^{2x} = ln5\)
Solve for the real domain
\(e ^{2x} = ln(5)\)
if \(f(x) =g(x),\) then \(ln(f(x))= ln(g(x))\)
\(ln(e ^{2x} ) = ln(ln(5))\)
Solve : \(ln(e ^{2x} ) = ln(ln(5))\)
use the logarithmic definition :
\(ln(e^{f(x)} ) = f(x)\)
\(ln(e^{2x} ) = 2x\)
\(2x=ln(ln(5))\)
Divide both sides by 2 :
\(\frac{2x}{2} = \frac{ln(ln(5))}{2}\)
\(x= \frac{ln(ln(5))}{2}\)
hope this helps!
