How do you solve #4^(x+5)=0.5#?

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Answer 1 · 2016-04-05T13:12:47

#x=-11/2#

#4^(x+5)=0.5#

First apply logarithms because #color (blue) (a=b => lna=lnb, if a,b>0)#

#(x+5)ln4=ln(0.5)#

#(x+5)ln(2^2)=ln(2^-1)#

#(x+5)*2*ln(2)=-ln(2)#

ln(2) is a constant, so you can divide the expression by it

#(x+5)*2=-1#

#2x+10=-1#

#2x=-11#

#x=-11/2#