How do you solve for x in $2^(3x-2)*3^(2x-3)=36^(x-1)$ ?

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Answer 1 · 2016-04-12T16:36:52

$x=0,63$

$2^(3x-2)*3^(2x-3)=36^(x-1)$

$2^(3x)/2^2*3^(2x)/3^3=36^x/36$

$2^(3x)/2^2*3^(2x)/3^3=6^(2x)/6^2$

$(2^(3x)*3^(2x))/6^(2x)=(2^2*3^3)/6^2$

$(2^(3x)*3^(2x))/6^(2x)=3$

$(2^(3x)*3^(2x))/(2*3)^(2x)=3$

$(2^(3x)*cancel(3^(2x)))/(2^(2x)*cancel(3^(2x)))=3$

$2^(3x-2x)=3$

$2^x=3$

$log 2^x=log 3$
$x*log2=log3$

$x=log2 /log3$

$log2=0,3010299957$
$log3=0,4771212547$

$x=0,63$

How do you solve for x in 2^(3x-2)*3^(2x-3)=36^(x-1) 2^(3x-2)*3^(2x-3)=36^(x-1)?