How do you solve "ln(8x-1) = 2x"?

Question

The equation is constructed from e^2x = 8x-1

ln(8x-1) = 2x

Answer 1

Graph the two equations:

y = ln(8x-1)
y = 2x

The point of intersection of the two graphs is the answer.

Answer 2

No algebraic way to solve it
Let´s take
f(x) = e^2x-8x+1
At -infinity the limit is + infinity
At +infinity the limit is +infinity
The function is always continuos
f´(x)= 2e^2x -8
f´(x) = 0 e^2x= 4 and x= 1/2* ln4
the sign of f´is -------1/2ln4+++++++
so we have a minimum at x= 1/2ln4
f(1/2ln4)= 4 -4ln4+1 = 5-4ln4 <0
so there are two roots.
One is between 1/2ln4 and 1 and the other between 0 an 1/2ln4

Answer 3

Multiply In times 8X=In8x
Multiply In times -1=In-1
Divide both sides of the equation by 2
X=(In8x)+In-1
Factor out In
X=In(8x)+1