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

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

$2x+10=-1$

$2x=-11$

$x=-11/2$

How do you solve 4^(x+5)=0.54^(x+5)=0.5?