How do you solve the equation ${log}_{2} (y + 2) - {log}_{2} (y - 2) = 1$ $log_2(y+2)-log_2(y-2)=1$ ?

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Answer 1 · 2016-10-23T02:12:27

$y = 6$ $y=6$

${log}_{2} (y + 2) - {log}_{2} (y - 2) = 1$ $log_2(y+2)-log_2(y-2)=1$

We can combine the two terms on the left side:

${log}_{2} (\frac{y + 2}{y - 2}) = 1$ $log_2((y+2)/(y-2))=1$

We can now deal with the log:

$2^{{log}_{2} (\frac{y + 2}{y - 2})} = 2^{1}$ $2^(log_2((y+2)/(y-2)))=2^1$

$\frac{y + 2}{y - 2} = 2$ $(y+2)/(y-2)=2$

$(y + 2) = 2 (y - 2)$ $(y+2)=2(y-2)$

$y + 2 = 2 y - 4$ $y+2=2y-4$

$6 = y$ $6=y$

And let's check it (which can also help get more comfortable with log operations):

${log}_{2} (y + 2) - {log}_{2} (y - 2) = 1$ $log_2(y+2)-log_2(y-2)=1$

${log}_{2} (6 + 2) - {log}_{2} (6 - 2) = 1$ $log_2(6+2)-log_2(6-2)=1$

${log}_{2} (8) - {log}_{2} (4) = 1$ $log_2(8)-log_2(4)=1$

$3 - 2 = 1$ $3-2=1$

$1 = 1$ $1=1$

How do you solve the equation log2(y+2)−log2(y−2)=1log_2(y+2)-log_2(y-2)=1?