A friend of mine asked me this 'seemingly" easy math question. But to my surprise, I am unable to solve it despite my university education level. So, the question goes like this.
$(5x+1)^2 = 0.2\sqrt{5^x}$
As you can, the left hand side constitutes a quadratic expression while the right hand is a term of base $5$ to the power of $x$ (exponential term). The problem arises when I log both sides and got stuck. Let me show you.
$(5x+1)^2 = \frac{1}{5}*5^\frac{x}{2}$
$\ln5(5x+1)^2 = \ln 5^\frac{x}{2}$
$2*\ln5(5x+1) = \frac{x}{2}*\ln 5$
and I'm stuck here, unsure of how to deal with the $x$ term in the natural logarithm at the LHS. Please give me some advice or help me out. I have spent hours thinking about this question.
*I have tried using wolfram alpha to solve for this equation. Apparently, there are three $x$ values which are $x = -0.279916, x = - 0.114587$ and $x = 12.2717$ and I suspect the answer to be the positive $x$ value because log term must be positive.
I believe some sort method exist to solve this kind of question: LHS = quadratic , RHS = exponent. However, I also realize for this question, the number $5$ seems to be quite important, maybe there's some special way to simplify without using log. I don't know
Update: Newton-Raphson iteration method is the way to solve this (aside from graphical method and finding the intersection point). Apparently, this method is useful for solving complicaed nonlinear equation and finding the roots via derivative. The formula is as shown,
you can write $$(5x+1)^2=5^{x/2-1}$$ and then $$2\ln(5x+1)=(\frac{x}{2}-1)\ln(5)$$ and then use a numerical method