For the logarithms: #2^(x+1) = 3^x#, how do you solve for x?

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Answer 1 · 2015-10-04T03:37:12

#x = ln(2)/(ln(3) - ln(2))#

Take the log of the two sides

#ln(2^(x+1)) = ln(3^x)#

Pass the exponents to the front of the log

#(x+1)ln(2) = xln(3)#

Expand the left side

#xln(2) + ln(2) = xln(3)#

Isolate x

#ln(2) = xln(3)-xln(2)#

Put x in evidence

#ln(2) = x(ln(3)-ln(2))#

Pass that dividing

#x = ln(2)/(ln(3) - ln(2))#

Or, if you want, you can use log properties to change that into other logs, like

#x = ln(2)/(ln(3/2))# or #x = log_(3/2)(2)#