How do you solve #4^ { x + 3} = 3^ { - 3x }#?

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Answer 1 · 2017-11-14T12:24:50

#x=-0.8883#

Taking log on both sides of #4^(x+3)=3^(-3x)#, we get

#(x+3)log4=-3xlog3#

or #x(log4+3log3)=-3log4#

or #x=-(3log4)/(log4+3log3)#

= #-(3xx0.6021)/(0.6021+3xx0.4771)#

= #-1.8063/2.0334#

= #-0.8883#