How do you solve $2^(x+1)=3^x$ ?

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Answer 1 · 2015-02-17T16:52:18

I would write it as:
$2^x*2^1=3^x$
$(2^x)/(3^x)=1/2$
$(2/3)^x=1/2$
take the log of both sides:
$ln(2/3)^x=ln(1/2)$
$xln(2/3)=ln(1/2)$
$x=1.71$

Answer 2 · 2015-02-19T17:28:48

The answer is: $x=ln2/(ln3-ln2)$ .

$2^(x+1)=3^xrArrln2^(x+1)=ln3^xrArr(x+1)ln2=xln3rArr$

$xln2+ln2=xln3rArrx(ln2-ln3)=-ln2rArrx=-ln2/(ln2-ln3)=ln2/(ln3-ln2)$ .

How do you solve 2^(x+1)=3^x2^(x+1)=3^x?