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)#.