How do you solve $2^{3 - x} = 565$ $2^(3-x) = 565$ ?

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Answer 1 · 2016-07-21T14:29:23

$x = 3 - \frac{log 565}{log 2} = - 6.142107057$ $x=3-(log 565)/(log 2)=-6.142107057$

We start from the given equation

$2^{3 - x} = 565$ $2^(3-x)=565$

Take the logarithm of both sides

${log 2}^{3 - x} = log 565$ $log 2^(3-x)=log 565$

$(3 - x) \cdot log 2 = log 565$ $(3-x)*log 2=log 565$

$3 - x = \frac{log 565}{log 2}$ $3-x=log 565/log 2$

$3 - \frac{log 565}{log 2} = x$ $3-log 565/log 2=x$

and

$x = 3 - \frac{log 565}{log 2} = - 6.142107057$ $x=3-(log 565)/(log 2)=-6.142107057$

God bless....I hope the explanation is useful

How do you solve 2^(3-x) = 56523−x=5652^(3-x) = 565?