Answer: \(\displaystyle AB=\frac{\sqrt{101}}{5}\), so \(AB\approx2.01\).

Use the laws of logarithms and exponentials first, then check that any logarithm arguments are positive where required.

At the point \(A\), \(x=0\), so

\(y=5e^0-3=2.\)

Thus

\(A=(0,2).\)

Differentiate the curve:

\(\frac{dy}{dx}=10e^{2x}.\)

At \(x=0\), the gradient is

\(10e^0=10.\)

So the tangent at \(A\) has equation

\(y-2=10(x-0).\)

Hence

\(y=10x+2.\)

At \(B\), the tangent meets the \(x\)-axis, so \(y=0\):

\(0=10x+2.\)

Therefore

\(x=-\frac15,\)

and

\(B=\left(-\frac15,0\right).\)

Now use the distance formula:

\(AB=\sqrt{\left(0+\frac15\right)^2+(2-0)^2}.\)

So

\(AB=\sqrt{\frac{1}{25}+4} =\sqrt{\frac{101}{25}} =\frac{\sqrt{101}}{5}.\)

Therefore

\(AB\approx2.01.\)