How do you solve $2^(x+1)=5^(1-2x)$ ?

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Answer 1 · 2015-02-05T15:06:44

The answer is: $x=(ln5-ln2)/(ln2+2ln5)$ .

The only way to find the solutions of this equation is to pass both members as arguments of the logarithmics function, better in base $e$ .

So:

$2^(x+1)=5^(1-2x)rArrln2^(x+1)=ln5^(1-2x)rArr$

$(x+1)ln2=(1-2x)ln5rArrxln2+ln2=ln5-2xln5rArr$

$xln2+2xln5=ln5-ln2rArrx(ln2+2ln5)=ln5-ln2rArr$

$x=(ln5-ln2)/(ln2+2ln5)$ .

How do you solve 2^(x+1)=5^(1-2x)2^(x+1)=5^(1-2x)?