How do you solve $3^{2 x + 1} = 5^{200}$ $3^(2x+1)=5^200$ ?

Question

2838 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-30T15:10:02

$\Rightarrow x = \frac{200 log 5 - log 3}{2 log 3}$ $=>x=(200log5-log3)/(2log3)$

$3^{2 x + 1} = 5^{200}$ $3^(2x+1)=5^200$

Taking log on both sides we have

$log (3^{2 x + 1}) = log (5^{200})$ $log(3^(2x+1))=log(5^200)$

$\Rightarrow (2 x + 1) log (3) = 200 log (5)$ $=>(2x+1)log(3)=200log(5)$

$\Rightarrow 2 x log 3 + log 3 = 200 log 5$ $=>2xlog3+log3=200log5$

$\Rightarrow 2 x log 3 = 200 log 5 - log 3$ $=>2xlog3=200log5-log3$

$\Rightarrow x = \frac{200 log 5 - log 3}{2 log 3}$ $=>x=(200log5-log3)/(2log3)$

How do you solve 32x+1=52003^(2x+1)=5^200?