Exponential equation question

Question

I've been trying to get my head around the following equation, I am doing this because I like doing maths, a friend of mine gave me these equations to tackle, I am no expert on exponential equations. $$ 0.2^{2x-2} + 5 \cdot 0.04^{x+1} = 126 $$ I seem to be unable to get it, after trying with all I know, this one seems a bit too difficult and I would like to ask you for some help / explanation as to how I would tackle this.

Thanks in advance

Answer 1

Hint:

note that $0.04=0.2^2$. so you can write the equation as: $$ 0.04^{x-1}+5\times0.04^{x+1}=126 $$ Can you do from this?

Answer 2

Notice that $0.04=0.2^2=5^{-2}$. Hence your equation is

$$0.04^{x-1}+5\cdot0.04^{x+1}=0.04^x(0.04^{-1}+5\cdot0.04)=126$$

or

$$5^{-2x}=5.$$

Answer 3

Note that $0.2 = \tfrac{1}{5}$ and $0.04 = \tfrac{1}{25} = \left(\tfrac{1}{5}\right)^2$ and use this to transform everything to an easier base: $$\left(\tfrac{1}{5}\right)^{2x-2}+5\left(\tfrac{1}{25}\right)^{x+1} =\left(\tfrac{1}{5}\right)^{2x-2}+\left(\tfrac{1}{5}\right)^{2x+1} =5^{-2x+2}+5^{-2x-1}=5^3 \cdot 5^{-2x-1}+5^{-2x-1}$$ Now you have: $$5^3 \cdot 5^{-2x-1}+5^{-2x-1} = 126 \iff 5^{-2x-1} \left( 125+1 \right) = 126$$ So...