Answer to Question #90454 in Statistics and Probability for eustace kusangwa

Question

5^(2x+1)+4=21*5^x

Answer 1

Substitution.

Let $5^x=t, t>0.$ Then $5^{2x+1}=5t^2$

$$5t^2+4=21t$$

$$5t^2-21t+4=0$$

$$D=(-21)^2-4(5)(4)=361$$

$$t={21\pm\sqrt {361} \over 2(5)}$$

$$t_1={21-19 \over 10}={1 \over 5}, \ \ t_2={21+19 \over 10}=4$$

$$5^x={1 \over 5}=>x=-1$$

$$5^x=4=>x=log_5(4)={2ln2 \over ln5}$$

$$\Big\{-1,\ log_5(4)\Big\} \ \ \ or \ \ \ \Big\{-1,\ {2ln2 \over ln5}\Big\}$$