How do you solve $8^{y} = 4^{2 x + 3}$ $8^y = 4^(2x+3)$ and $log 2 y = log 2 x + 4$ $log2y=log2x+4$ ?

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How do you solve $8^{y} = 4^{2 x + 3}$ $8^y = 4^(2x+3)$ and $log 2 y = log 2 x + 4$ $log2y=log2x+4$ ?

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Answer 1 · 2016-04-30T04:41:52

The first equation becomes

$8^{y} = 4^{2 x + 3} \Rightarrow {(2^{3})}^{y} = 2^{2 \cdot (2 x + 3)} \Rightarrow 3 \cdot y = 4 x + 6$ $8^y=4^(2x+3)=>(2^3)^y=2^(2*(2x+3))=>3*y=4x+6$

The second equation becomes

$log 2 y = log 2 x + 4 \Rightarrow log (\frac{2 y}{2 x}) = 4 \Rightarrow log (\frac{y}{x}) = 4 \Rightarrow \frac{y}{x} = 10^{4} \Rightarrow y = 10^{4} \cdot x$ $log2y=log2x+4=>log((2y)/(2x))=4=>log(y/x)=4=> y/x=10^4=>y=10^4*x$

Now the system of equations become

$3 y = 4 x + 6$ $3y=4x+6$
$y = 10^{4} \cdot x$ $y=10^4*x$

which has solutions

$x = \frac{3}{14998}$ $x = 3/14998$ , $y = \frac{15000}{7499}$ $y = 15000/7499$

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How do you solve $8^{y} = 4^{2 x + 3}$ $8^y = 4^(2x+3)$ and $log 2 y = log 2 x + 4$ $log2y=log2x+4$ ?

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