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

2 Answers
Gió
Feb 17, 2015

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#

Massimiliano
Feb 19, 2015

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)#.

Related questions
See all questions in Natural Logs
Impact of this question
1266 views around the world
You can reuse this answer
Creative Commons License