How do you solve # 3^(2x) = 4^(x+1)#?

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Answer 1 · 2015-09-18T15:54:16

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

Take the log of both sides
#log(3^(2x)) = log(4^(x+1))#

Remember the property that #log(m^n) = nlog(m)#

#2xlog(3) = (x+1)log(4)#

Expand

#2xlog(3) = xlog(4) + log(4)#

Put everything with an x to the same side

#2xlog(3) - xlog(4) = log(4)#

Put x on evidence, and put that 2 back in the log

#x(2log(3) - log(4)) = log(4)#

Isolate x

#x = log(4)/(2log(3) - log(4))#

Remember that #4 = 2^2#, so #log(4) = 2log(2)

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

Put 2 in evidence and cancel it

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